cadaVR anatomy by Mozaik Education

Equations, inequalities

Problems leading to linear equations, inequalities

Look for logical connections in the text, and choose the variables.
Formulate an equation.
Solve it.
Check the solution according to the text.
State the answer.

Example 1

Solve the following equation on the set of real numbers:

\latex{7 \times \left(x+3\right)-5 \times \left(x-1\right)=3 \times \left(x+4\right)+6x}.

Solution

First expand the parentheses on both sides, then combine the homogeneous terms:

\latex{7x+21-5x+5=3x+12+6x},
\latex{2x+26=9x+12}.

After subtracting \latex{12} and \latex{2x} from both sides, we get

\latex{14 = 7x},
\latex{2=x}.

Substitution to the original equation gives \latex{30} on both sides, therefore the solution of the problem is \latex{x = 2}.

Example 2

The \latex{85\%} of grapes is water, while only \latex{18\%} of the raisins sold in shops is water. How much grape is needed to produce \latex{500\,g} of raisins?

Solution

Let us denote the appropriate mass in kilograms by \latex{x}. Part of the water is lost during the drying process, thus we should compute the amount of dry matter within the fruit in both cases. In fresh grapes: \latex{x \times 0.15\,kg}, in the \latex{500\,g} of raisins sold in shops: \latex{0.5 \times 0.82}.

The amount of “dry material” is the same in both cases: \latex{x \times 0.15 = 0.5 \times 0.82}, from which

\latex{x=\frac{0.5 \times 0.82}{0.15}},
\latex{x=2.733}.

Correctness of the result can be easily checked by substituting back into the first equation. Therefore \latex{2.73\,kg} of grapes are needed for \latex{500\,g} of raisins.

Example 3

Is there any solution of the following equation among the integers?

\latex{\frac{x+5}{x+3}-\frac{x-4}{x-3}=\frac{5x-2}{x^2-9}}.

Solution

The base set is \latex{\Z}. Because of the denominators, \latex{x + 3 \ne 0,\space x-3 \ne 0} and \latex{x^2-9 \ne 0}. All criteria are satisfied if \latex{x \ne-3} and \latex{x \ne 3}.

Multiply both sides by \latex{x^2-9}, the common denominator of the fractions, then expand the parentheses and combine the homogeneous terms:

\latex{\left(x+5\right)\times\left(x-3\right)-\left(x-4\right)\times\left(x+3\right)=5x-2},
\latex{x^2-3x+5x-15-x^2-3x+4x+12=5x-2},
\latex{3x-3=5x-2}.

Subtract \latex{3x} then add \latex{2} to both sides. Finally, divide both by \latex{2}:

\latex{-1=2x},
\latex{x=-\frac{1}{2}}.

The resulting solution is not element of the base set, therefore the equation does not have a solution on the set of integers.

Example 4

At \latex{8} am a motorist departs from Szeged to Debrecen, a city \latex{224\,km} away, with a speed of \latex{50\,km/h}. At \latex{8:40} am another motorist leaves from Debrecen to Szeged with a speed of \latex{60\,km/h}. When will they meet? Which motorist covers more distance by then?

Solution

Let us denote the travel time of the motorist departing from Szeged by \latex{x}. The travel time of the one departing from Debrecen is \latex{40} minutes, or \latex{\frac{2}{3}} hours, shorter. Construct a table using the data available:

motorist departing
from Szeged

motorist departing
from Debrecen

Velocity (\latex{ km/h })

Time (\latex{ h })

Distance covered (\latex{ km })

\latex{50}

\latex{x}

\latex{50x}

\latex{60}

\latex{x-\frac{2}{3}}

\latex{60 \times \left(x-\frac{2}{3}\right)}

The sum of the distance covered by the two motorists until they met is the distance of the two cities, therefore the following equation can be obtained:

\latex{50x+60\times \left(x-\frac{2}{3}\right)=224},

\latex{50x+60x-40=224},

\latex{110x-40=224},

\latex{110x = 264},

\latex{x = 2.4}

They meet \latex{2.4} hours, that is, \latex{2} hours and \latex{24} minutes after the first motorist's departure.

Let us check the result: the motorist leaving Szeged covers \latex{120} km during \latex{2.4} hours, the other one covers \latex{104} km during \latex{1} hour and \latex{44} minutes, the sum of the two distances covered is indeed \latex{224} km.

The motorists meet at \latex{10:24} am and until then, the one leaving from Szeged covers more distance.

Example 5

Solve the following equation on the set of real numbers:

\latex{|2x-3|-4=|x+5|}.

Solution

First we have to understand the expressions with absolute value in the equation:

\latex{|2x-3|=\begin{cases}2x-3\text{, if }x\ge\frac{3}{2}\text{,}\\-2x+3\text{, if }x\lt\frac{3}{2}\text{,}\end{cases}\quad} and \latex{\quad|x+5|=\begin{cases}x+5\text{, if }x\ge-5\text{,}\\-5-x\text{, if }x\lt-5\text{,}\end{cases}}

We can divide the real number line into three parts according to the domains above as seen on Figure 6.

Case I: If \latex{x\lt-5}, then the
equation is

\latex{-2x+3-4=-x-5},
\latex{-2x-1=-x-5},
\latex{4=x}.

Case II: If \latex{-5\le x \lt \frac{3}{2}}, then the
equation is

Case III: If \latex{\frac{3}{2}\le x}, then the
equation is

\latex{-2x+3-4=x+5},
\latex{-2x-1=x+5},
\latex{-6=3x},
\latex{-2=x}.

\latex{2x-3-4=x+5},
\latex{2x-7=x+5},
\latex{x=12}.

Since \latex{x=4} is not an element
of the base set, it is not a solu-
tion.

\latex{x=-2} is an element of the
base set, and it is possible to
check that it is indeed a solu-
tion.

The resulting root satisfies the
criterion of the case III and it
can be verified by checking it
that it is indeed a solution.

Thus the solutions of the equation are (Figure 7):

\latex{x_1=-2\quad} and \latex{\quad x_2=12}

Note: The equation can be solved graphically as well. This way, it is the intersection points of the functions corresponding to the two sides of the equation that are needed to be found. (Figure 8)

\latex{x}

\latex{y}

\latex{y=|x+5|}

\latex{y=|2x-3|-4}

\latex{12}

\latex{-2}

Figure 8

Example 6

Solve the following equation on the set of real numbers:

\latex{\sqrt{3x+1}=4};

\latex{\sqrt{4x+3}+\sqrt{-1-x}=2}.

Solution (a)

First find the domain of the equation. Since the expression taken the root of cannot be negative, \latex{3x + 1\ge 0}, that is (Figure 9),

\latex{x\ge-\frac{1}{3}}

Both sides are non-negative, in such a case taking the square of both sides is an equivalent transformation. Doing this we get

\latex{3x + 1 = 16},
\latex{3x = 15},
\latex{x=5}.

The resulting root is indeed a solution as it is an element of the domain, and the same can be seen after a check: \latex{\sqrt{3 \times 5+1}=\sqrt{16}=4.} (Figure 10)

Solution (b)

Inspection of the domain: \latex{4x + 3 \ge 0} and \latex{–1 – x \ge 0}, that is,

\latex{x \ge -\frac{3}{4}\quad} and \latex{\quad x\le-1}.

There is no common element in the set of solutions for the two inequalities. Since the domain is the empty set, the equation does not have any solution. (Figure 11)

Example 7

Solve the following equation on the set of real numbers:

\latex{\sqrt{7-x}-\sqrt{-1-2}=2}.

Solution

Inspection of the domain (Figure 12): \latex{7-x\ge0} and \latex{-1-x\ge 0}. The solutions of these inequalities are \latex{7 \ge x} and \latex{-1\ge x}. The intersection of the two sets is \latex{-1\ge x}.

Rearrange the equation such that both sides are non-negative:

\latex{\sqrt{7-x}=2+\sqrt{-1-x}}.

By taking the square of both sides, we have:

\latex{7-x=4+4\times \sqrt{-1-x}-1-x}.

After rearrangement:

\latex{4=4\times\sqrt{-1-x}}.

After dividing by \latex{4} and taking the square:

\latex{1=\sqrt{-1-x}},
\latex{1=-1-x},
\latex{x=-2}.

The result is an element of the domain.

Check the value of the left hand side of the original equation:

\latex{\sqrt{7-\left(-2\right)}-\sqrt{-1-\left(-2\right)}=\sqrt{9}-\sqrt{1}=2}.

Therefore the only solution of the problem is \latex{x = –2}.

Example 8

Solve the following inequality on the set of real numbers:

\latex{3 \times \left(x-7\right)-5\times \left(2x-3\right)\gt7\times \left(3x-4\right)}.

Solution

After the expansion of the parentheses, let us combine the like terms:

\latex{x3-21-10x+15 \gt 21x-28},
\latex{-7x-6 \gt 21x-28}.

Rearrange the inequality:

\latex{-28x\gt-22},
\latex{x\lt\frac{22}{28}}.

The result, after simplification by \latex{2}:

\latex{x\lt\frac{11}{14}}.

Example 9

Find the natural numbers for which the following inequality holds:

\latex{\frac{3}{2} \times x -1-\frac{1}{5}\times \left(5x-4\right)\lt\frac{x+11}{10}}.

Solution

Find a common denominator and then multiply both sides:

\latex{\frac{15x}{10}-1-\frac{2\times \left(5x-4\right)}{10}\lt\frac{x+11}{10}},

\latex{15x-10-2\times\left(5x-4\right)\lt x+11},

Expand the parentheses and then rearrange the inequality:

\latex{15x-10-10x+8\lt x+11},

\latex{5x-2\lt x+11},

\latex{4x\lt 13},

\latex{x\lt\frac{13}{4}}.

Compare the result to the domain (Figure 13). Thus the natural numbers satisfying the inequality are \latex{\lbrace0; 1; 2; 3\rbrace}.

Example 10

For which values of parameter \latex{p} will the solution of the equation \latex{3 \times(x-2)=p \times \left(x+2\right)-1} be greater than \latex{1}?

Solution

Expand the parentheses and then solve the equation:

\latex{3 \times \left(x-2\right)=p \times \left(x+2\right)-1},
\latex{3 x-6=px+2p-1},
\latex{3 x-px=2p+5},
\latex{x\times \left(3-p\right) =2p+5}.

If \latex{p = 3}, then the equation does not have any solutions.

If \latex{p \ne 3}, then the solution is

\latex{x=\frac{2p+5}{3-p}}.

We are looking for those values of \latex{p} for which

\latex{\frac{2p+5}{3-p}\gt1}.

After rearranging and finding a common denominator:

\latex{\frac{2p+5}{3-p}-1\gt0},

\latex{\frac{2p+5-3+p}{3-p}-1\gt0},

\latex{\frac{3p+2}{3-p}-1\gt0},

A fraction is positive if the signs of its numerator and denominator are the same. There are two possible cases.

Case I: \latex{\quad\begin{gather*}3p+2 \gt0\text{,}\\p\gt-\frac{3}{2}\text{,}\end{gather*}\quad} and \latex{\quad\begin{gather*}3-p \gt0\text{,}\\3\gt p\text{.}\end{gather*}\quad}

Illustration can be seen in Figure 14. The intersection of the sets is

\latex{-\frac{2}{3}\lt p\lt 3}.

Case II: \latex{\quad\begin{gather*}3p+2 \lt0\text{,}\\p\lt-\frac{3}{2}\text{,}\end{gather*}\quad} and \latex{\quad\begin{gather*}3-p \lt0\text{,}\\3\lt p\text{.}\end{gather*}\quad}

Illustration can be seen in Figure 15. The two sets do not have any elements in common.

Therefore the solution of the original equation is greater than \latex{1} if

\latex{-\frac{2}{3}\lt p \lt 3}.

Example 11

Solve the following inequality on the set of real numbers:

\latex{\sqrt{2x-3}\lt2}.

Solution

The radical expression in the inequality is valid if \latex{2x-3 \ge 0}, that is

\latex{x \ge\frac{3}{2}}.

Since both sides are non-negative, taking the square of both sides does not reverse the relation between them: \latex{2x-3\lt4}, which means

\latex{x\lt\frac{7}{2}}.

Comparison with the domain shows that the solution is _{(Figure 16):}

\latex{\frac{3}{2}\le x \lt \frac{7}{2}}.

Problems leading to quadratic equations, inequalities

THEOREM: The quadratic equation \latex{ax^2+bx+c=0 \left(a\neq 0\right)} has

– two real roots,
\latex{x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}}, \latex{x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}},
if \latex{D=b^2-4ac \gt0};
– one real root, \latex{x=-\frac{b}{2a}}, if \latex{D=b^2-4ac=0};
– no real roots, if \latex{D=b^2-4ac\lt0};

DEFINITION: The form \latex{a \times \left(x-x_1\right)\times\left(x-x_2\right)=0\text{, }D \ge0} is called the zero product form of the quadratic equation.

The following relations hold between the roots and coefficients of the quadratic equation \latex{ax^2+bx+c=0 \quad\left(a \ne0\right)}:

\latex{x_1+x_2=-\frac{b}{a}\quad} and \latex{\quad x_1 \times x_2=\frac{c}{a}}.

These identities are called Vieta's formulae.

Example 12

The length of a football field is \latex{42\,m} more than its width. Its centre is \latex{62\,m} away from the corner flags. Find the size of the field.

Solution

Draw a picture. Connect the centre of the rectangle with one vertex and the midpoint of the longer side. Let \latex{x} denote the length of this second segment. The length of the other leg of the right angled triangle created is \latex{x+21}, according to the problem, while the length of the hypotenuse is \latex{62\,m}. (Figure 17)

Apply the Pithagorean theorem:

\latex{2^2+\left(x+21\right)^2=62^2}

After rearrangement:

\latex{2x^2+42x-3403=0},

Apply the quadratic formula:

\latex{x_{1,2}=\frac{-42 \pm\sqrt{42^2+4\times 2 \times 3403}}{4}}.

The results are

\latex{x_1=32.065\quad} and \latex{\quad x_2=-53.065}.

Naturally, only the positive root is a solution for the problem, thus the value in question is approximately \latex{32\,m}.

The football field is approximately \latex{ 64\, m } wide and \latex{ 106\, m } long.

Example 13

It takes \latex{2} hours longer for a mother to solve a specific jigsaw puzzle alone than it takes for her son. Together they can solve it in \latex{2} hours and \latex{24} minutes.

How long would it take for the son to solve the puzzle alone?

Solution

Let \latex{x} denote the time the boy would need to solve the puzzle alone, then the time for the mother to solve it alone is \latex{x+2} hours.

In an hour, the boy completes \latex{\frac{1}{x}} part of the puzzle, while the mother completes \latex{\frac{1}{x+2}} part.

When working together, their efforts add up, thus they can complete \latex{\frac{\frac{1}{12}}{5}=\frac{5}{12}} part of the puzzle in an hour.

The following equation holds:

\latex{\frac{1}{x}+\frac{1}{x+2}=\frac{5}{12}},

After multipication:

\latex{12 \times \left(x+2\right)+12x=5x \times \left(x+2\right)},

\latex{12 x+24+12x=5x^2+10x}.

After rearrangement:

\latex{x_{1,2}=\frac{14 \pm \sqrt{14^2+4 \times 5 \times 24}}{10}},

which leads to

\latex{x_1=4\quad} and \latex{\quad x_2=-\frac{6}{5}}.

Obviously, only the positive root can be a solution, and it can be checked that it is indeed correct.

Therefore the son can solve the puzzle in \latex{ 4 } hours, and the mother can finish in \latex{ 6 } hours.

Example 14

The parametric quadratic equation \latex{3x^2-5x+c=0} is given.

For which values of parameter \latex{c} are there any real solutions of the equation?
For which values of \latex{c} are there two negative solutions?
For which value of \latex{c} will the sum of the squares of the roots be equal to \latex{3}?
For which values of \latex{c} can the fraction \latex{\frac{3x^2-5x+c}{x^2+x-12}} be simplified?

Solution (a)

A quadratic equation has real solutions if its discriminant is nonnegative:

\latex{\begin{array}{r}\left(-5\right)^2-4 \times 3 \times c \ge0\text{,}\\25-12c \ge0\text{,}\end{array}}
\latex{c \le \frac{25}{12}}.

Thus there exists a real solution if \latex{c\le\frac{25}{12}}.

Solution (b)

According to the Vieta's formulae,

The discriminant of the quadratic equation \latex{ax^2+bx+c=0} is: \latex{D=b^2-4ac}. If \latex{D\gt0}, then there exists \latex{2} real solutions. If \latex{D=0}, then there exists \latex{1} real solution (\latex{2} coinciding roots). If \latex{D\lt0}, then there is no real solution.

\latex{x_1+x_2=\frac{5}{3}},

\latex{x_1 \times x_2=\frac{c}{3}},

Since the sum of the roots is positive, they cannot be both negative. Therefore there is no parameter value \latex{c} for which both roots are negative.

Solution (c)

Express the sum of the roots' squares using the sum and the product of the roots:

\latex{{x_1}^2+{x_2}^2=\left(x_1+x_2\right)^2-2\times x_1 \times x_2}.

Substitute the formulas from the previous part:

\latex{3=\frac{25}{9}-\frac{2c}{3}},

\latex{c=-\frac{1}{3}}.

Since \latex{-\frac{1}{3}\lt \frac{25}{12}}, there are indeed two solutions for the resulting value.

Therefore the sum of the roots' squares will be \latex{3} if \latex{c=-\frac{1}{3}}.

Solution (d)

The fraction is valid if

\latex{x^2+x-12 \ne 0}.

The solutions of this equation are

\latex{x \ne4 \quad} and \latex{\quad x \ne 3}.

The denominator can be factorized using the factored form:

\latex{x^2+x-12=\left(x+4\right)\times \left(x-3\right)}.

The fraction can be simplified if the numerator can be factorized such that one factor is either \latex{x+4} or \latex{x-3}. This is possible if either \latex{-4} or \latex{3} is a solution of the equation \latex{3x^2-5x+c=0}.

If \latex{x=-4}:

\latex{3 \times 16-5 \times\left(-4\right)+c=0},
\latex{c=-68}.

Then the equation and its solutions are:

\latex{3x^2-5x-68=0},

\latex{x_1=-4 \quad} and \latex{\quad x_2=\frac{17}{3}}.

In this case the fraction can be indeed simplified:

\latex{\frac{3x^2-5x-68}{x^2+x-12}=\frac{3\times\left(x+4\right)\times\left(x-\frac{17}{3}\right)}{\left(x+4\right)\times\left(x-3\right)}=\frac{3x-17}{x-3}}.

If \latex{x=3}:

\latex{\begin{align*}{r}3 \times 9-5 \times 3 +c =0,\\27-15+c=0, \\c=-12.\end{align*}}

The equation and its solutions are:

\latex{3x^2-5x-12=0},

\latex{x_1=3 \quad} and \latex{\quad x_2=-\frac{4}{3}}.

In this case the fraction can be simplified as well:

\latex{\frac{3x^2-5x-12}{x^2+x-12}=\frac{3\times\left(x-3\right)\times\left(x+\frac{4}{3}\right)}{\left(x+4\right)\times\left(x-3\right)}=\frac{3x+4}{x+4}}.

Therefore the fraction can be simplified if either

\latex{c=-68 \quad} or \latex{\quad c=-12}.

Example 15

Solve the following equation on the set of real numbers:

\latex{\sqrt{6+x}+\sqrt{4-x}=\sqrt{5x+1}}.

Solution

The equation is valid if the following three inequalities hold (Figure 18):

\latex{\begin{array}{c}6+x\ge0,\\\space\\x\ge-6,\end{array}\quad} and \latex{\quad\begin{array}{c}4-x\ge0,\\\space\\x\le 4,\end{array}\quad} and \latex{\quad\begin{array}{c}5x+1\ge0,\\\space\\x\ge-\frac{1}{5}.\end{array}}

The intersection of the three sets is

\latex{-\frac{1}{5}\le x \le4}.

Take the square of both sides:

\latex{6+x+2 \times \sqrt{\left(6+x\right) \times\left(4-x\right)}+4-x=5x+1}.

Combine and rearrange:

\latex{2 \times \sqrt{\left(6+x\right) \times\left(4-x\right)}=5x-9}.

Taking the square of both sides once more gives

\latex{4 \times \left(-x^2-2x+24\right)=25x^2-90x+81}.

Reducing to \latex{ 0 } gives

\latex{0=29x^2-82x-15}.

Application of the quadratic formula gives us the solutions:

\latex{x_1=3\quad} and \latex{\quad x_2=-\frac{5}{29}}.

Both numbers are elements of the domain, but substituting \latex{x_2} into the equation before taking the square the second time gives a negative value on the right hand side, therefore \latex{x_2} is not a solution. Substitution of \latex{x_1} shows that it is indeed a solution.

Example 16

Solve the following equation on the set of real numbers:

\latex{|2x^2-11x+5|=x+5}.

Solution

Inspect the sign of the expression inside the absolute value.

Solutions of the quadratic equation \latex{2x^2-11x+5=0}:

\latex{x_1\quad} and \latex{\quad x_2=\frac{1}{2}}

In accordance to this,

\latex{|2x^2-11x+5|=\begin{cases}\quad x^2-11x+5\text{1 if }x\le\frac{1}{2}\text{ or }x\ge5\text{,}\\\space\\-2x^2+11x-5\text{, if }\frac{1}{2}\lt x \lt 5\text{.}\end{cases}}

Case I: If \latex{x \le \frac{1}{2}} or \latex{x \ge5}, then the equation is

\latex{2x^2-11x+5=x+5},

\latex{2x^2-12x=0}.

Factorization shows that the roots are

\latex{x_1=0\quad} and \latex{\quad x_2=6.}

Both are elements of the domain of this case, therefore they are both solutions.

Case II: If \latex{\frac{1}{2}\lt x \lt 5}, then the equation is

\latex{-2x^2+11x-5=x+5},

\latex{2x^2+10x+10=0}.

Application of the quadratic formula gives

\latex{x_1=\frac{5+\sqrt{5}}{2}\quad} and \latex{\quad x_2=\frac{5-\sqrt{5}}{2}}.

Again, these numbers are elements of the domain of the case, therefore they are also solutions.

The solutions of the equation are

\latex{x_1=0}; \latex{\quad x_2=6};

\latex{x_3=\frac{5+\sqrt{5}}{2}}; \latex{\quad x_4=\frac{5-\sqrt{5}}{2}}.

Example 17

Solve the following inequality on the set of real numbers:

\latex{\frac{x^2+3x-10}{2x^2-5x-3}\lt2}.

Solution

The fraction is valid if its denominator is not zero: \latex{2x^2-5x-3 \ne 0}.

Apply the quadratic formula:

\latex{x \ne 3 \quad} and \latex{\quad x \ne -\frac{1}{2}}.

Rearrange the inequality:

\latex{\frac{x^2+3x-10}{x2^2-5x-3}-2\lt0}.

Find the common denominator:

\latex{\begin{align*}\frac{x^2+3x-10-4x^2+10x+6}{2x^2-5x-3}\lt0\text{,}\\\\\frac{-3x^2+13x-4}{2x^2-5x-3}\lt0\text{.}\end{align*}}

Now find the zero points of the numerator as well. The solutions of the equation \latex{-3x^2 +13x-4 = 0} are:

\latex{x_1 = 4 \quad} and \latex{\quad x_2 = \frac{1}{3}}.

Using the zero points of the numerator and denominator the two quadratic functions with their images being the expressions corresponding to the numerator and denominator, respectively, can be displayed. (Figure 19)

The fraction is negative if the signs of its numerator and denominator are different. The intervals in question can be found using the figure. The solution is

\latex{x\lt-\frac{1}{2}\quad} or \latex{\quad\frac{1}{3}\lt x\lt3\quad} or \latex{\quad x\gt4}.

Example 18

Solve the following inequality on the set of real numbers:

\latex{\sqrt{x^2-3x-4}\ge x-1}.

Solution

The formula is valid if \latex{x^2-3x-4\ge0}.

The roots of the quadratic expression are \latex{x_1 = 4} and \latex{x_2 =-1}. Thus the solution of the criterion is (Figure 20):

\latex{x \le -1 \quad} and \latex{\quad x \ge 4}.

The left hand side of the inequality is non-negative.

Case I: If the right hand side is non-negative as well, \latex{x-1\ge0}, that is, \latex{x\ge1}, then taking the square of both sides does not change the relation between the sides:

\latex{x^2-3x-4 \ge x^2-2x+1}

The solution of this inequality is \latex{x \le-5}.

Comparison to the criterion \latex{x \ge1} shows that there is no solution on this set. (Figure 21)

Case II: If the right hand side is negative, \latex{x-1\lt0}, that is, \latex{x\lt1}, then the inequality holds for every \latex{x} in the domain. (Figure 22)

The criteria provided by the domain and by this case are satisfied if \latex{x\le-1}. Therefore the solution of the inequality is \latex{x\le-1}.

Exponential and logarithmic equations and inequalities

Example 19

Solve the following equations on the set of real numbers:

\latex{\sqrt[3]{49}=\sqrt{7^x}};

\latex{8 \times 4^x=\sqrt[5]{16}}.

Solution (a)

Since \latex{49 = 7^2}, both sides can be written as a power of \latex{7}:

\latex{7^{\frac{2}{3}}=7^{\frac{x}{2}}}.

The exponential function of base \latex{7} is strictly increasing (Figure 23), thus

\latex{\frac{2}{3}=\frac{x}{2}}, from which \latex{x=\frac{4}{3}}.

Correctness of the solution can be seen after a check.

Solution (b)

Express both sides as a power of \latex{2}:

\latex{2^3\times \left(2^2\right)^x=\left(2^4\right)^{\frac{1}{5}}}.

After combination:

\latex{2^{3+2x}=2^{\frac{4}{5}}}.

The exponential function of base \latex{ 2 } is strictly increasing, thus

\latex{3+2x=\frac{4}{5}}, from which \latex{x=-\frac{11}{10}}.

The solution can be checked by computation.

Example 20

Solve the following equations on the set of real numbers:

\latex{3 \times 5^{x+1}-4 \times 5^{x}=55};

\latex{4^{\frac{x}{2}-1}-16^{\frac{x}{4}-1}=2^{x+1}+31}.

Solution (a)

Using \latex{5^{x+1}=5 \times 5^x}, the equation can be formed as:

\latex{3 \times 5 \times 5^x-4 \times 5^x = 55},

\latex{15 \times 5^x-4 \times 5^x = 55},

\latex{11 \times 5^x = 55},

\latex{5^x = 5},

\latex{x= 1},

Substituting into the left hand side gives \latex{55} as well. Thus the solution of the equation is \latex{x = 1}.

Solution (b)

Express the base of every power as a power of \latex{2}: \latex{4 = 2^2, 16 = 2^4}.

\latex{\left(2^2\right)^{\frac{x}{2}+1}-\left(2^4\right)^{\frac{x}{4}-1}=2^{x+1}+31},

\latex{2^{x+2}-2^{x-4}=2^{x+1}+31},

Move the powers of \latex{2} on one side and transform the exponents:

\latex{2^x \times 2^2-\frac{2^x}{2^4}-2^x\times2=31},

\latex{2^x\times \left(4-\frac{1}{16}-2\right)=31},

\latex{2^x\times\frac{31}{16}=31},

\latex{2^x=16},

\latex{x=4}.

Substituting gives \latex{63} for both sides. Therefore the solution of the equation is \latex{x=4}.

Example 21

Solve the following equations on the set of real numbers:

\latex{9^{x+\frac{1}{2}}-28 \times 3^{x}+9=0};

\latex{4^{x+3}-2^{x+4}-8=0}.

Solution (a)

Transform the first term of the left hand side:

\latex{9^{\frac{1}{2}}\times9^x - 28\times 3^x +9=0},

\latex{3\times9^x - 28 \times 3^x+9=0}.

Since \latex{9^x=\left(3^x\right)^2}, it is worth introducing a new variable: \latex{y = 3^x}.

The equation using the new variable:

\latex{3y^2-28y+9=0}.

Apply the quadratic formula:

\latex{y_1=9 \quad} and \latex{\quad y_2=\frac{1}{3}}.

Substituting gives

\latex{3^x=9 \quad} and \latex{\quad 3^x=\frac{1}{3}}.

The solutions are:

\latex{x=2 \quad} and \latex{\quad x=-1}.

Substitution into the original equation verifies that these are both solutions.

Solution (b)

Since \latex{4^{x+3}=\left(2^2\right)^{x+3}=\left(2^{x+3}\right)^2}, the second term should be transformed accordingly:

\latex{2^{x+4}=2 \times 2^{x+3}}.

The equation is

\latex{\left(2^{x+3}\right)^2-2 \times 2^{x+3}-8=0}.

Introduction of a new variable \latex{y=2^{x+3}} changes the equation into

\latex{y^2-2y-8=0}.

Solve using the quadratic formula:

\latex{y_1=4 \quad} and \latex{\quad y_2=-2}.

Since a power of \latex{2} cannot be negative, the second one does not give a solution.

Substituting into the first one gives

\latex{2^{x+3}=4},

\latex{x+3=2},

\latex{x=-1}.

Substitution into the original equation shows that this is indeed a solution of the equation.

Example 22

Solve the following inequalities on the set of real numbers:

\latex{25^{\frac{3}{2}}\gt5^{|x|-1}};

\latex{9^x-6 \times 3^x-27\gt0}.

Solution (a)

Transform the left hand side: \latex{5^3\gt5^{|x|-1}}.

The exponential function of base \latex{5} is strictly increasing, thus

\latex{3\gt|x|-1},

\latex{4\gt |x|}.

The solution of this is \latex{-4 \lt x \lt 4}.

Solution (b)

Introduce a new variable: \latex{y = 3x}. Using this, the equation becomes \latex{y^2-6y-27\gt0}.

Applying the quadratic formula on the quadratic equation \latex{y^2-6y-27=0} gives

\latex{y_1=9\quad} and \latex{\quad y_2=-3}.

The solution of the inequality is: \latex{y\lt -3} or \latex{y \gt 9}.

A power of \latex{3} cannot be negative, therefore there is no \latex{x} belonging to the first inequality.

The second inequality gives \latex{3^x \gt 9}. The exponential function of base \latex{3} is strictly increasing, thus \latex{x \gt 2}.

Example 23

Solve the following equations on the set of real numbers:

\latex{\log_3\left(4x-3\right)^2=2 \times \log_3\left(4x-3\right)};
\latex{\log\left(3x-5\right)-\log\left(5x-3\right)-\log\left(3-5x\right)=\log\left(3x+5\right)};
\latex{\log_5 |2x-7|=\log_5\left(-2-5x\right)}.

Solution (a)

The equation is valid if \latex{\left(4x-3\right)^2\gt0} and \latex{4x-3\gt0} holds.

If \latex{x\gt\frac{3}{4}}, then both criteria are satisfied.

According to the identity for the logarithm of a power, for every element of the domain,

\latex{\log_3\left(4x-3\right)^2=2\times \log_3\left(4x-3\right)}.

Therefore the solution of the equation is \latex{x \gt \frac{3}{4}}.

Solution (b)

To find the domain of the equation one has to solve the following inequalities:

\latex{3x-5\gt0\quad} and \latex{\quad 5x-3\gt0}

and \latex{\quad 3-5x\gt0\quad} and \latex{\quad 3x+5\gt0}.

Their solutions are (Figure 24):

\latex{x\gt \frac{5}{3} \quad}and \latex{\quad x\gt \frac{3}{5}\quad} and \latex{\quad x\lt \frac{5}{3}\quad} and \latex{\quad x \gt -\frac{5}{3}}.

The four inequalities do not have any solutions in common, therefore the domain of the equation is the empty set. Thus the equation does not have any solutions.

Solution (c)

The equation is valid if \latex{|2x-7| \gt0} and \latex{-2-5x\gt0}.

The first criterion is satisfied if \latex{x \ne \frac{7}{2}} while the second one is satisfied if \latex{x \lt -\frac{2}{5}} Both criteria are satisfied for \latex{x \lt -\frac{2}{5}}.

\latex{\log_5|2x-7|= \log_5\left(-2-5x\right)}.

Since the logarithmic function of base \latex{5} is strictly increasing (Figure 25), the logarithms of two expressions are equal if and only if the expressions are equal as well.

In this case,

\latex{|2x-7|=-2-5x}.

We have to distinguish two cases because of the absolute value.

Case I: If \latex{x\gt\frac{7}{2}};

\latex{2x-7=-2-5x},
\latex{7x=5},
\latex{x=\frac{5}{7}}.

Case II: If \latex{x\lt\frac{7}{2}};

\latex{-2x+7=-2-5x},
\latex{3x=-9},
\latex{x=-3}.

Does not satisfy the criterion,
thus not a solution.

Satisfies the condition and an
element of the domain as well.
Substituting gives \latex{\log_513} for
both sides.

The solution of the equation is \latex{x = -3}.

Example 24

Solve the following equations on the set of real numbers:

\latex{\log\left(x+3\right)-\log\left(x-4\right)=\log 2 + \log\left(x-1\right)};
\latex{\log 3 + \log \sqrt{6x-5}=\log\left(8x-6\right)};
\latex{\log x +\log \left(x^2-14\right)=3 \times \log\left(x-2\right)}.

Solution (a)

The equation is valid if

\latex{x+3\gt 0 \quad} and \latex{\quad x-4\gt 0\quad} and \latex{\quad x-1\gt0}.

The solutions of these are:

\latex{x\gt-3 \quad} and \latex{\quad x\gt 4\quad} and \latex{\quad x\gt1}.

The common solution is (Figure 26): \latex{x \gt 4}.

Using the identities for the sum and the difference of logarithms, both sides can be replaced as seen below:

\latex{\log \frac{x+3}{x-4}= \log \left(2x-2\right)}.

Since the logarithmic function is strictly increasing,

\latex{\log\frac{x+3}{x-4}=\log\left(2x-2\right)},

\latex{x+3=\left(x-4\right) \times \left(2x-2\right)},

\latex{x+3=2x^2-10x+8},

\latex{0=2x^2-11x+5}.

The solutions of the quadratic equation are

\latex{x_1 =5\quad} and \latex{\quad x_2=\frac{1}{2}}.

Only the first one is an element of the domain of the equation.

Substituting gives \latex{\log 8} for both sides, thus the solution is \latex{x=5}.

Solution (b)

The equation is valid if

\latex{6x-5\gt0\quad} and \latex{\quad 8x-6\gt0}.

Their solutions are, respectively,

\latex{x \gt \frac{5}{6}\quad} and \latex{\quad x \gt \frac{3}{4}}.

Both inequalities are satisfied for \latex{x \gt \frac{5}{6}}.

The left hand side can be transformed using the identity for the sum of logarithms:

\latex{\log \left(3 \times \sqrt{6x-5}\right)=\log \left(8x-6\right)}.

Since the logarithmic function is strictly increasing,

\latex{3 \times \sqrt{6x-5}=8x-6}.

Rearranging after taking the square of both sides gives

\latex{9 \times \left(6x-5\right)=\left(8x-6\right)^2},

\latex{54x-45=64x^2-96x+36},

\latex{0=64x^2-150x+81}.

Using the quadratic formula gives

\latex{x_1 = \frac{3}{2}\quad} and \latex{\quad x_2 = \frac{27}{32}}.

Both results are greater than \latex{\frac{5}{6}}. Substituting gives \latex{\lg6} for both sides in case of \latex{x_1}, and \latex{\lg\frac{3}{4}} in case of \latex{x_2}.

Therefore the solutions of the equation are

\latex{x_1 = \frac{3}{2}\quad} and \latex{\quad x_2 = \frac{27}{32}}.

Solution (c)

The equation is valid if

\latex{x^2-14\gt0 \quad} and \latex{\quad x-2\gt0}.

Their solutions are, respectively,

\latex{|x|\gt \sqrt{14} \quad} and \latex{\quad x\gt2}.

Since \latex{\sqrt{14}\gt2}, the domain of the equation is (Figure 27): \latex{x\gt \sqrt{14}}.

Using the identities of the logarithm for both sides gives

\latex{\log\left[ x \times\left(x^2-14\right)\right]=\log\left(x-2\right)^3}.

Since the logarithmic function is strictly increasing:

\latex{x \times \left(x^2-14\right)=\left(x-2\right)^3}.

Expanding the parentheses, then sorting leads to:

\latex{x^3-14x=x^3-6x^2+12x-8},

\latex{6x^2-26x+8=0}.

Divide by \latex{2}:

\latex{3x^3-13x+4=0}.

Roots can be obtained using the quadratic formula:

\latex{x_1=4 \quad} and \latex{\quad x_2=\frac{1}{3}}.

Only \latex{x_1} is an element of the domain. Substituting gives \latex{\log 8} for both sides, that is, the solution of the equation is \latex{x=4}.

Example 25

Solve the following inequalities on the set of real numbers:

\latex{\log_4\left(3x-1\right)-\log_4\left(x+2\right)\lt\log_4 2};
\latex{\log_{0.7}5 + \log_{0.7}\left(2x-1\right)\gt \log_{0.7} \left(3x+9\right)}.

Solution (a)

The inequality is valid only if

\latex{3x-1\gt 0 \quad} and \latex{\quad 2+x\gt 0}

These criteria give \latex{x\gt\frac{1}{3}} and \latex{x \gt -2} as solutions. The common solution is: \latex{x\gt\frac{1}{3}}.
(Figure 28)

The left hand side can be transformed using the identities of the logarithm:

\latex{\log_4\frac{3x-1}{x+2}\lt\log_42}.

As the logarithmic function of base \latex{4} is strictly increasing,

\latex{\frac{3x-1}{x+2}\lt2}.

Rearrange to one side:

\latex{\frac{3x-1}{x+2}-2\lt0}.

Find the common denominator:

\latex{\frac{3x-1}{x+2}-\frac{2x+4}{x+2}\lt0},

\latex{\frac{x-5}{x+2}\lt0}.

On the domain of the equation we have \latex{x+2\gt0}, therefore the fraction is negative if \latex{x-5\lt0}, that is, \latex{x\lt5}. (Figure 29)

Comparison with the domain shows that the solution of the inequality is

\latex{\frac{1}{3}\lt x \lt 5}.

Solution (b)

The inequality is valid only if

The equation is valid if

\latex{2x-1\gt0 \quad} and \latex{\quad 3x+9\gt0}.

Their respective solutions are \latex{x\gt\frac{1}{2}} and \latex{x\gt-3}. The common solution is (Figure 30): \latex{x\gt\frac{1}{2}}.

The left hand side can be transformed using the identities of the logarithm:

\latex{\log_{0.7}\left[5 \times\left(2x-1\right)\right]\gt\log_{0.7}\left(3x+9\right)}.

Since \latex{0.7 \lt1}, the logarithmic function of base \latex{0.7} is strictly decreasing (Figure 31):

\latex{5 \times \left(2x-1\right)\lt3x+9}.

After expanding, its solution is

\latex{10x-5\lt3x+9},

\latex{7x\lt14},

\latex{x\lt2}.

The solution of the inequality in accordance with the domain is (Figure 32):

\latex{\frac{1}{2}\lt x\lt2}.

Trigonometric equations, inequalities

\latex{\sin}

\latex{\frac{1}{2}}

\latex{\cos}

\latex{\tan}

\latex{\cot}

\latex{\frac{\sqrt{3}}{2}}

\latex{30°}

\latex{45°}

\latex{60°}

\latex{\frac{\sqrt{2}}{2}}

\latex{\frac{\sqrt{3}}{2}}

\latex{\frac{\sqrt{3}}{3}}

\latex{\sqrt{3}}

\latex{\frac{\sqrt{2}}{2}}

\latex{1}

\latex{\frac{1}{2}}

\latex{\sqrt{3}}

\latex{\frac{\sqrt{3}}{3}}

Example 26

Solve the following equation on the set of real numbers:

\latex{\cos^2x=\frac{3}{4}}.

Solution

Taking the square root of both sides yields the equation \latex{|\cos x|=\frac{\sqrt{3}}{2}}.

According to the absolute value we have to solve the following two equations:

\latex{\cos x =\frac{\sqrt{3}}{2} \quad} or \latex{\quad x=-\frac{\sqrt{3}}{2}}.

In both cases, draw a picture (Figure 33).

\latex{\frac{\sqrt{3}}{2}}

\latex{-\frac{\sqrt{3}}{2}}

\latex{\frac{\sqrt{3}}{2}}

\latex{-\frac{\sqrt{3}}{2}}

\latex{y}

\latex{x}

Figure 33

Figure 34

Using the diagrams, the solutions can be determined:

\latex{x_1 =\frac{\pi}{6}+k\times2\pi}, \latex{\quad x_2 =\frac{\pi}{6}+l\times2\pi},

\latex{x_3 =\frac{5\pi}{6}+m\times2\pi}, \latex{\quad x_4 =-\frac{5\pi}{6}+n \times2\pi},

where \latex{k}, \latex{l}, \latex{m} and \latex{n} are integers.

Uniting the diagrams makes it possible to unite the solutions as well. (Figure 34)

\latex{x=\pm\frac{\pi}{6}+k\times\pi}, where \latex{k \in \Z}.

The correctness of these results can be checked.

Example 27

Solve the following equation on the set of real numbers:

\latex{\tan\left(2x-\pi\right)=-1}.

Solution

By the definition of the tangent, \latex{\tan a=-1} if (Figure 35)

\latex{\alpha=-\frac{\pi}{4}+l\times\pi}, \latex{\left(l \in \Z\right)}.

Therefore,

\latex{2x-\pi=-\frac{\pi}{4}+l\times\pi},

\latex{2x=\frac{3\pi}{4}+l\times\pi},

\latex{x=\frac{3\pi}{8}+l\times\frac{\pi}{2}}, \latex{l \in \Z}.

The correctness of the result can be checked.

Example 28

Solve the following equation on the set of real numbers:

\latex{\cos\left(3x-\frac{\pi}{2}\right)=\cos\left(x+\pi\right)}.

Solution I

The cosines of two angles are equal if the angles are either equal or opposite, aside from the period (Figure 36).

Case I:

Case II:

\latex{3x-\frac{\pi}{2}=x+\pi+k \times 2 \pi},
\latex{2x=\frac{3 \pi}{2}+k \times 2 \pi},
\latex{x=\frac{3 \pi}{4}+k \times \pi} \latex{\quad\left(k \in \Z\right)}.

\latex{3x-\frac{\pi}{2}=-x-\pi+l \times 2 \pi},
\latex{4x=-\frac{\pi}{2}+l \times 2 \pi},
\latex{x=-\frac{\pi}{8}+l \times \frac{\pi}{2}} \latex{\quad\left(l \in \Z\right)}.

The correctness of the results can be checked.

Solution II

The solutions of the equation \latex{\cos \alpha=\cos \beta} can also be found using the graph of the function \latex{f(x) = \cos x} (Figure 37):

\latex{\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + k \times 2 \pi\quad} or \latex{\quad\alpha = -\beta+l\times 2 \pi \quad \left(k,l \in \Z\right)}.

\latex{y=\cos \alpha}

\latex{y=\cos x}

\latex{x}

\latex{y}

Figure 37

From here the computation can be finished similarly to Solution I.

*Solution III

After sorting to one side, the difference can be factorized:

\latex{\cos \alpha- \cos \beta = -2 \times\sin\frac{\alpha + \beta}{2}\times\sin \frac{\alpha-\beta}{2}}.

Using this, the original equation can be transformed into

\latex{\cos \left(3x-\frac{\pi}{2}\right) - \cos \left(x+\pi\right)=0},

from which it follows that

\latex{-2 \times \sin\frac{3x-\frac{\pi}{2}+x+\pi}{2}\times \sin\frac{3x-\frac{\pi}{2}-x-\pi}{2}=0}.

Sort the equation:

\latex{-2 \times \sin \frac{4x+\frac{\pi}{2}}{2}\times \sin \frac{2x-\frac{3\pi}{2}}{2}=0},

\latex{-2 \times \sin \left(2x+\frac{\pi}{4}\right)\times \sin \left(x-\frac{3\pi}{4}\right)=0}.

There are two possible cases:

\latex{\sin\left(2x+\frac{\pi}{4}\right)=0},\latex{\qquad\qquad} or \latex{\qquad\qquad}\latex{\sin\left(x-\frac{3\pi}{4}\right)=0},

\latex{2x+\frac{\pi}{4}=k \times \pi},\latex{\qquad\qquad}\latex{\qquad\qquad}\latex{x-\frac{3\pi}{4}=l \times \pi},

\latex{x=-\frac{\pi}{8}+k \times \frac{\pi}{2} \space \left( k \in \Z\right)},\latex{\qquad\qquad}\latex{\qquad\qquad}\latex{x=\frac{3\pi}{4}+l\times\pi \space \left(l \in \Z\right)}.

Example 29

Solve the following equation on the set of real numbers:

\latex{\cos 2x+3 \times \sin x =2}.

Solution

Use the identity for \latex{\cos 2x} on the left hand side:

\latex{1-2 \times \sin^2x+3\times\sin x=2}.

This is a quadratic equation for \latex{\sin x}:

\latex{2 \times sin^2x-3 \times\sin x+1=0}.

Apply the quadratic formula:

\latex{\sin x =1\quad} or \latex{\quad\sin x = \frac{1}{2}}.

Create a diagram (Figure 38), and use it to find the solutions:

\latex{x_1=\frac{\pi}{2}+k\times2\pi}, \latex{\quad x_2=\frac{\pi}{6}+l \times2\pi}, \latex{\quad x_3=\frac{5\pi}{6}+m\times 2\pi\qquad\left(k,l,m, \in \Z\right)}.

The correctness of the solutions can be checked.

Example 30

Find the real numbers for which \latex{\tan2\alpha=\cot\alpha}.

Solution

Use the identity for \latex{\tan 2a} and use that \latex{\tan a \times \cot a = 1}.

\latex{\frac{2 \times \tan\alpha}{1-\tan^2\alpha}=\frac{1}{\tan\alpha}}.

It can be seen easily using this form, that the equation is valid if \latex{|\tan \alpha| \ne 1} and \latex{\tan \alpha \ne 0}, that is, if

\latex{\alpha \ne \pm \frac{\pi}{4}+k\pi}; \latex{\alpha \ne l\pi}; \latex{\left(k,l \in \Z\right)}.

After multiplication,

\latex{2 \times \tan^2\alpha=1-\tan^2\alpha},

\latex{3 \times \tan^2\alpha=1},

\latex{\tan^2\alpha=\frac{1}{3}},

\latex{|\tan\alpha|=\frac{1}{\sqrt{3}}}.

This can be satisfied in two ways:

\latex{\tan \alpha=\frac{\sqrt{3}}{3}},\latex{\qquad\qquad} or \latex{\qquad\qquad}\latex{\tan\alpha=-\frac{\sqrt{3}}{3}},

\latex{\alpha=\frac{\pi}{6}+m\times\pi \quad \left(m\in\Z\right)},\latex{\qquad\qquad}\latex{\qquad\qquad}\latex{\alpha=-\frac{\pi}{6}+n\times\pi \quad \left(n\in\Z\right)}.

Example 31

Solve the following equation on the set of real numbers:

\latex{\cos x-\sqrt{3}\times \sin x =1}.

Solution

If both sides are divided by \latex{2}:

\latex{\frac{1}{2}\times \cos x-\frac{\sqrt{3}}{2}\times \sin x = \frac{1}{2}},

then the sum of the squares of the coefficients:

\latex{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2=1}.

Thus the coefficients can be viewed as trigonometric functions of an angle.

Let \latex{\frac{1}{2}=\cos \alpha} and \latex{\frac{\sqrt{3}}{2}=\sin \alpha}, then \latex{\alpha=\frac{\pi}{3}} is a possible angle. (Figure 39)

Then the equation is

\latex{\cos\frac{\pi}{3}\times \cos x- \sin\frac{\pi}{3}\times\sin x=\frac{1}{2}}.

The expression on the left hand side can be transformed using the addition formula:

\latex{\cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}}.

The solutions are

\latex{\frac{\Pi }{3}+x=\frac{\Pi }{3}+k\times 2\Pi ,} or \latex{\frac{\Pi }{3}+x=\frac{\Pi }{3}+l\times 2\Pi ,}
\latex{x=k\times 2\Pi (k\in \Z)}, \latex{x=-\frac{2\Pi }{3}+l\times 2\Pi (l\in \Z).}

Since the transformations were equivalent ones, the results are indeed solutions of the original equation.

Note: The equation can be solved by taking the square of both sides as well, but then at the end one has to check for false roots.

Example 32

Solve the following equation on the set of real numbers:

\latex{\cos 3x\geq \frac{\sqrt{2} }{2}.}

Solution I

Plot the possible values for \latex{3x} in the unit circle.

Figure 40 shows that there is equality at \latex{3x=\frac{\Pi }{4}} and \latex{3x=-\frac{\Pi }{4}}.

The solution of the inequality:

\latex{-\frac{\Pi }{4}+k\times 2\Pi \leq 3x\leq \frac{\Pi }{4}+k\times 2\Pi .}

Therefore the result is

\latex{-\frac{\Pi }{12}+k\times \frac{2\Pi }{3}\leq x\leq \frac{x}{12}+k\times \frac{2\Pi }{3} (k\in \Z).}

Solution II

Plot the functions \latex{f(x)=\cos x} and \latex{g(x)=\frac{\sqrt{2} }{2}} in one coordinate system. (Figure 41)

\latex{y=\frac{\sqrt{2} }{2}}

\latex{-\frac{\Pi }{2}}

\latex{-\frac{\Pi }{4}}

\latex{\frac{\Pi }{4}}

\latex{\frac{\Pi }{2}}

\latex{\Pi}

\latex{\frac{3\Pi }{2}}

\latex{y=\cos x}

\latex{ x }

Figure 41

The solution of the inequality can be found using the graph in Figure 41:

\latex{-\frac{\Pi }{4}+k\times 2\Pi \leq 3x\leq \frac{\Pi }{4}+k\times 2,}

\latex{-\frac{\Pi }{12}+k\times \frac{2\Pi }{3}\leq x\leq \frac{\Pi }{12}+k\times \frac{2\Pi }{3} (k\in \Z).}

Example 33

Solve the following equation on the set of real numbers:

\latex{\sin 2x\gt \sqrt{3}\times \cos x.}

Solution

Substitute for \latex{\sin 2x} on the left hand side.

\latex{2\times \sin x\times \cos x=\sqrt{3}\times \cos x.}

Rearrange to one side and factorize:

\latex{\cos x\times (2\times \sin x-\sqrt{3} )\gt 0.}

There are two ways for a product to be positive: either both factors are positive (Case I) or they are negative (Case II).

CASE I: \latex{\cos x\gt 0} and \latex{2\times \sin x-\sqrt{3}\gt 0,} that is \latex{\sin x\gt \frac{\sqrt{3} }{2}.}

Illustrate the conditions on the unit circle.

Condition I:

Condition II:

Intersection
of conditions:

\latex{ y }

\latex{ x }

Figure 42

Their solutions are (Figure 42):

Condition I:\latex{-\frac{\Pi }{2}+k\times 2\Pi \lt x\lt \frac{\Pi }{2}+k\times 2\Pi ,}

Condition II:\latex{\frac{\Pi }{3}+l\times 2\Pi \lt x\lt \frac{2\Pi }{3}+l\times 2\Pi (k,l\in \Z) .}

The intersection of the two intervals (Figure 42):

\latex{\frac{\Pi }{3}+m\times 2\Pi \lt x\lt \frac{\Pi }{2}+m\times 2\Pi (m\in \Z) .}

CASE II: \latex{\cos x\lt 0} and \latex{2\times \sin x-\sqrt{3}\lt 0,} that is \latex{\sin x\lt \frac{\sqrt{3} }{2}.}

Again, illustrate using the unit circle.

Condition I:

Condition II:

Intersection
of conditions:

\latex{ y }

\latex{ x }

Figure 43

The solution of the two inequalities (Figure 43):

Condition I: \latex{\frac{\Pi }{2}+k\times 2\Pi \lt x\lt \frac{3\Pi }{2}+k\times 2\Pi ,}

Condition II: \latex{\frac{2\Pi }{3}+l\times 2\Pi \lt x\lt \frac{7\Pi }{3}+l\times 2\Pi (k\in \Z).}

The intersection of the intervals (Figure 43):

\latex{\frac{2\Pi }{3}+m\times 2\Pi \lt x\lt \frac{3\Pi }{2}+m\times 2\Pi (m\in \Z).}

To summarize the two cases, the solution of the inequality is (Figure 44):

\latex{\frac{\Pi }{3}+m\times 2\Pi \lt x\lt \frac{\Pi }{2}+m\times 2\Pi}

or \latex{\frac{2\Pi }{3}+m\times 2\Pi \lt x\lt \frac{3\Pi }{2}+m\times 2\Pi (m\in \Z).}

Exercises

{{exercise_number}}. How much do we need to mix from a \latex{ 40 }% and a \latex{ 80 }% solution to end up with \latex{ 10 } liters of \latex{ 50 }% solution?

{{exercise_number}}. The last digit of a \latex{ 3 }-digit number is \latex{ 3 }. If we place this digit at the end of the number instead of the start, we get a number smaller by \latex{ 162 }. What is this \latex{ 3 }-digit number?

{{exercise_number}}. During a bicycle trip we have completed the \latex{\frac{2}{3}} part of the distance planned for that day until noon. After the lunch break we covered one third of the distance planned before \latex{ 5 } pm and there was \latex{ 24\, km } remaining. How many kilometers did we cover that day?

{{exercise_number}}. Tommy completed the \latex{ 40 }% of the book given as required reading. On the next day he read \latex{ 135 } pages, after which there was only \latex{ 30 }% of the book left. How many pages are there in the book?

{{exercise_number}}. It takes \latex{ 2.5 } hours by car to travel from Szeged to Budapest. It takes \latex{ 3 } hours by a lorry which covers \latex{ 12 } less \latex{ km } per hour. What is the distance of the two cities?

{{exercise_number}}. A pool can be filled up using two taps. If only the tap with higher performance is opened, then the pool fills up in \latex{ 3 } hours, if only the tap with smaller performance, then it fills up in \latex{ 6 } hours. At what time should both taps be opened, at latest, if the pool has to be filled up by \latex{ 6 } am?

{{exercise_number}}. Examine the fraction \latex{\frac{8}{n-7}.} For what values of the natural number \latex{ n } will the fraction be

natural number;

negative real number;

greater than \latex{\frac{1}{2}?}

{{exercise_number}}. The pool in the public bath of Mórahalom is \latex{ 12\, m } longer than it is wide. The depth of water is \latex{ 2\, m }. Determine the size of the pool if \latex{1,386\,m^{3}} water is needed to fill it up.

{{exercise_number}}. Two workers have to produce \latex{ 400 } parts each. One of them produces \latex{ 5 } more in an hour than the other. How long do they work and how many parts do each of them produce in an hour if it take them a total of \latex{ 36 } hours to finish?

{{exercise_number}}. A French merchant sells an item for \latex{ 39 } euros. How much did he buy it for if he sells it for as many percent more as many euros it cost him?

{{exercise_number}}. For which real parameter \latex{ p } is there exactly one solution for the following equation?

\latex{(2+\log _{2}p )\times x^{2}+(6\times \log _{2}p )\times x+4\times \log _{2}p+1=0.}

{{exercise_number}}. Determine the value of the parameter \latex{ p } such that the following equation has a real solution.

\latex{\sqrt{x^{2}-8x-65 }+\sqrt{x^{2}-169 }+\sqrt{x^{2}+(p+14)\times +5p+19 }=0.}

{{exercise_number}}. Solve the following equations on the set of real numbers.

\latex{125^{x-3}=\sqrt[5]{25^{x} };}

\latex{4^{|2x+6|-|x-9|}=8;}

\latex{9^{x+\sqrt{x^{2}+2 } }-4\times 3^{x-1+\sqrt{x^{2+2} } }=69.}

{{exercise_number}}. Solve the following equations on the set of real numbers.

\latex{\log (2x-3)-\log (x+1)=\log \frac{1}{2};}

\latex{\log _{5}(2x+5)+\log _{5}x=\log _{5}(15-2x);}

\latex{\log 2+\log (4^{x}+9 )=1+\log (2^{x}+1 ).}

{{exercise_number}}. Solve the following equation on the set of natural numbers.

\latex{\log _{2}3+\log _{2^{2} }3^{2}+...+\log _{2^{n} }3^{n}=\log _{2}81.}

{{exercise_number}}. Solve the following inequalities on the set of real numbers.

\latex{\frac{3x-2}{2x-3}\lt 2;}

\latex{\frac{|x+3|+x}{x+2}\gt 1;}

\latex{(x+1)\times \sqrt{4-x^{2} }\leq 0;}

\latex{\sqrt{13^{x}-5 }\leq \sqrt{2\times (13^{x}+12 )}-\sqrt{13^{x}+5 }.}

{{exercise_number}}. Solve the following equations on the set of real numbers.

\latex{2\times \sin ^{2}x=1;}

\latex{\sin \left(2x-\frac{\Pi }{3} \right)=\sin \frac{x}{2};}

\latex{\sin 2x-3\times \cos x=0;}

\latex{\sin x+\cos x=1.}

{{exercise_number}}. Solve the following inequalities on the set of real numbers.

\latex{\sin (x-\Pi )\geq -\frac{\sqrt{3} }{2};}

\latex{\cos 2x+\cos x\leq 0.}

Mathematics 12.

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