Precalculus

Trigonometry

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