# Precalculus

## Trigonometry # Solving Trigonometric Equations Using Boundedness of Functions

We know that the sine and cosine functions are bounded:

$\vert{\sin x}\vert \le 1,\; \vert{\cos x}\vert \le 1$

The functions secant and cosecant are also bounded:

$\vert{\sec x}\vert \ge 1,\; \vert{\csc x}\vert \ge 1$

The boundedness property of functions can be used in solving certain trigonometric equations.

Consider an example. Let it be required to solve the equation

$\sin 2x + \sin 3x = 2.$

Since both sines do not exceed 1, this equation can only be satisfied if both functions are simultaneously equal to 1:

$\sin 2x + \sin 3x = 2, \Rightarrow \left\{ \begin{array}{l} \sin 2x = 1\\ \sin 3x = 1 \end{array} \right..$

Solving this system, we get

$\left\{ \begin{array}{l} 2x = \frac{\pi}{2} + 2\pi n\\ 3x = \frac{\pi}{2} + 2\pi k \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x = \frac{\pi}{4} + \pi n\\ x = \frac{\pi}{6} + \frac{2\pi k}{3} \end{array} \right., \;n,k \in \mathbb{Z}.$

The general solution of the system satisfies the equality

$\frac{\pi}{4} + \pi n = \frac{\pi}{6} + \frac{2\pi k}{3}.$

Multiply both sides by 12 and divide by $$\pi.$$ This yields:

$3 + 12n = 2 + 8k, \;\text{ or }\; 8k = 12n + 1.$

This equality is impossible because there is an even number on the left side and an odd number on the right side. Therefore, this equation has no solutions: $$x \in \varnothing.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the trigonometric equation

$\sec^2x = 1- x^2.$

### Example 2

Prove that the equation

$3\sin x + 4\cos x = x^2 - 2x +7$

has no solutions.

### Example 3

Solve the equation

$\sin 7x + \cos 2x = -2.$

### Example 4

Solve the equation

$\cos 2x\cos 3x = -1.$

### Example 1.

Solve the trigonometric equation

$\sec^2x = 1- x^2.$

Solution.

The function secant squared on the left side is bounded from below. Its minimum value is 1. The quadratic function on the right side, on the contrary, has a maximum value equal to 1. So the solution exists only if

$\left\{ \begin{array}{l} \sec^2x = 1\\ 1-x^2 = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} \cos^2x = 1\\ x^2 = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} \cos x = \pm 1\\ x = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x = \pi n, n \in \mathbb{Z}\\ x = 0 \end{array} \right..$

The system has only one solution $$x = 0$$ at $$n = 0$$.

### Example 2.

Prove that the equation

$3\sin x + 4\cos x = x^2 - 2x +7$

has no solutions.

Solution.

Using the $$R-$$method, we can convert a linear combination of sine and cosine into one sine function:

$a\sin \omega t + b\cos \omega t = c\sin \left({\omega t + \varphi}\right),$

where the amplitude $$c$$ and the phase $$\varphi$$ are given by

$c = \sqrt{a^2 + b^2},\;\;\varphi = \arctan\frac{b}{a}.$

In our case we get

$3\sin x + 4\cos x = \sqrt{3^2 + 4^2}\sin\left({x + \arctan\frac{4}{3}}\right) = 5\sin\left({x + \arctan\frac{4}{3}}\right).$

It can be seen that the left side of the equation does not exceed 5.

Consider the quadratic function on the right side. Let's represent it in the form

$x^2 - 2x + 7 = \left({x^2 - 2x + 1}\right) + 6 = \left({x - 1}\right)^2 + 6.$

This function has a minimum value of 6 when $$x = 1$$. Therefore, the original equation has no solutions.

### Example 3.

Solve the equation

$\sin 7x + \cos 2x = -2.$

Solution.

This equality holds only if both terms on the left side are equal to $$-1.$$ Hence,

$\sin 7x + \cos 2x = -2, \Rightarrow \left\{ \begin{array}{l} \sin 7x = -1\\ \cos 2x = -1 \end{array} \right..$

Solve each equation:

$\left\{ \begin{array}{l} 7x = \frac{3\pi}{2} + 2\pi n\\ 2x = \pi + 2\pi k \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x = \frac{3\pi}{14} + \frac{2\pi n}{7}\\ x = \frac{\pi}{2} + \pi k \end{array} \right., \;n,k \in \mathbb{Z}.$

The general solution of the system is determined by the condition

$\frac{3\pi}{14} + \frac{2\pi n}{7} = \frac{\pi}{2} + \pi k.$

Multiply both sides by 14 and divide by $$\pi:$$

$3 + 4n = 7 + 14k, \Rightarrow 2n = 7k + 2,$

where $$n$$ and $$k$$ are arbitrary integers.

$$2n$$ on the left side is an even number. To have an even number on the right side, $$k$$ must also be an even number. Let $$k = 2m, m \in \mathbb{Z}.$$ Then

$2n = 7k + 2 = 14m + 2, \Rightarrow n = 7m + 1.$

Now we can write the values of $$x:$$

$x = \frac{\pi}{2} + \pi k = \frac{\pi}{2} + 2\pi m,\;m \in \mathbb{Z}.$

The same answer is obtained if we express $$x$$ from the first equation of the system:

$x = \frac{3\pi}{14} + \frac{2\pi n}{7} = \frac{3\pi}{14} + \frac{2\pi \left({7m + 1}\right)}{7} = \frac{3\pi + 28\pi m + 4\pi}{14} = \frac{7\pi + 28\pi m}{14} = \frac{\pi}{2} + 2\pi m,\;m \in \mathbb{Z}.$

### Example 4.

Solve the equation

$\cos 2x\cos 3x = -1.$

Solution.

Perhaps this equation can be solved using trigonometric identities, but we will use the boundedness property of cosine.

It is clear that two options may arise here - when the first function is positive and the second is negative and vice versa.

#### Case 1.

$\left\{ \begin{array}{l} \cos 2x = 1\\ \cos 3x = -1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} 2x = 2\pi n\\ 3x = \pi + 2\pi k \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x = \pi n\\ x = \frac{\pi}{3} + \frac{2\pi k}{3} \end{array} \right.,\;n,k \in \mathbb{Z}.$

The solution is given by

$\pi n = \frac{\pi}{3} + \frac{2\pi k}{3}, \Rightarrow 3n = 1 + 2k.$

$$1 + 2k$$ always takes an odd value. Therefore $$n$$ must also be odd: $$n = 2m + 1,$$ where $$m \in \mathbb{Z}.$$ It follows that

$x = \pi n = 2\pi m + \pi,\;m \in \mathbb{Z}.$

#### Case 2.

$\left\{ \begin{array}{l} \cos 2x = -1\\ \cos 3x = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} 2x = \pi + 2\pi n\\ 3x = 2\pi k \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x = \frac{\pi}{2} + \pi n\\ x = \frac{2\pi k}{3} \end{array} \right.,\;n,k \in \mathbb{Z}.$

Here the solution is determined by the condition

$\frac{\pi}{2} + \pi n = \frac{2\pi k}{3}, \Rightarrow 3 + 6n = 4k,\;n,k \in \mathbb{Z}.$

The last equality is impossible for any $$n$$ and $$k$$ since the left side is odd and the right side is even. Therefore, in this case: $$x \in \varnothing.$$

Answer: $$x = 2\pi m + \pi,\;m \in \mathbb{Z}.$$