# Solving Trigonometric Equations Using Boundedness of Functions

We know that the sine and cosine functions are bounded:

The functions secant and cosecant are also bounded:

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

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

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

Solving this system, we get

The general solution of the system satisfies the equality

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

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

### Example 1.

Solve the trigonometric equation

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

The system has only one solution \(x = 0\) at \(n = 0\).

### Example 2.

Prove that the equation

has no solutions.

Solution.

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

where the amplitude \(c\) and the phase \(\varphi\) are given by

In our case we get

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

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

### Example 3.

Solve the equation

Solution.

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

Solve each equation:

The general solution of the system is determined by the condition

Multiply both sides by 14 and divide by \(\pi:\)

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

Now we can write the values of \(x:\)

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

### Example 4.

Solve the equation

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.

The solution is given by

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

#### Case 2.

Here the solution is determined by the condition

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