General Solution of Trigonometric Equations

A trigonometric equation is any equation that contains trigonometric functions. We have learned about trigonometric identities, which are satisfied for every value of the involved angles whereas, trigonometric equations are satisfied only for some values (finite or infinite in number) of the angles. In real life, we use trigonometric equations for making roof inclination, installing ceramic tiles, and building and navigating directions.

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This Story also Contains

1. Trigonometric Equations
2. Solution of Trigonometric Equation
3. General Solution of some Standard Equations
4. Important Points to remember while solving trigonometric equations
5. Summary
6. Solved Examples Based on Solution of Trigonometric Equations

In this article, we will cover the concept of Solution of Trigonometric Equations. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of thirty-one questions have been asked on this concept, including one in 2016, two in 2017, one in 2019, one in 2020, seven in 2021, nine in 2022 and eight in 2023.

Trigonometric Equations

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Trigonometric equations are satisfied only for some values (finite or infinite in number) of the angles. A value of the unknown angle that satisfies the given trigonometric equation is called a solution or a root of the equation. For example, equation 2 sin x = 1 is satisfied by x = π/6 is the solution of the equation between o and π. The solutions of a trigonometric equation lying in the interval [0,π ) are called principal solutions.

We know that trigonometric ratios are periodic functions. Functions sin x, cos x, sec x, and cosec x are periodic with period 2π and functions tan x and cot x are periodic functions with period π. Therefore, solutions of trigonometric equations can be generalized with the help of the period of trigonometric functions.

What are Trigonometric Equations?

The linear equation $a x+b=0$ can be written as a trigonometric equation as

$
a \sin \theta+b=0
$

which is also sometimes written as

$
\sin \theta=\sin \alpha
$

The quadratic equation

$
a x^2+b x+c=0
$

as an example of a trigonometric equation is written as

$
a \cos ^2 \theta+b \cos \theta+c=0
$

But unlike normal solutions of equations, where the number of solutions is based on the degree of the variable, in trigonometric equations, the same value of the solution exists for different values of $\theta$.

For example,
$\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}=\sin \frac{5 \pi}{6}=\sin \frac{13 \pi}{6}$, and so on, as the values of the sine function repeat after every $2 \pi$ radians.
Examples of Trigonometric Equations

$
\begin{gathered}
\sin 2 x-\sin 4 x+\sin 6 x=0 \\
2 \cos ^2 x+3 \sin x=0 \\
\cos 4 x=\cos 2 x \\
\sin 2 x+\cos x=0 \\
\sec ^2 2 x=1-\tan 2 x
\end{gathered}
$

Solution of Trigonometric Equation

The value of an unknown angle that satisfies the given trigonometric equation is called a solution or root of the equation. For example, $2 \sin \theta=1$, clearly $\theta=30^{\circ}$ satisfies the equation; therefore, $30^{\circ}$ is a solution of the equation. Now trigonometric equation usually has infinite solutions due to the periodic nature of trigonometric functions. So this equation also has (360+30)^o,(720+30)^o,(-360+30)^o, and so on, as its solutions.

Some important proofs.

1. Prove that for any real numbers x and y, sin x = sin y implies x = nπ + (-1)ⁿy, where n ∈ Z

Proof: If $\sin x=\sin y$, then

$
\begin{gathered}
\sin x-\sin y=0 \\
\Rightarrow 2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=0 \\
\Rightarrow \cos \left(\frac{x+y}{2}\right)=0 \quad \text { or } \quad \sin \left(\frac{x-y}{2}\right)=0 \\
\Rightarrow \frac{x+y}{2}=\frac{(2 n+1) \pi}{2} \quad \text { or } \quad \frac{x-y}{2}=n \pi, \quad \text { where } n \in \mathbb{Z}
\end{gathered}
$

[Because $\sin A=0$ implies $A=n \pi$ and $\cos A=0$ implies $A=\frac{(2 n+1) \pi}{2}$, where $n \in \mathbb{Z}$ ]

$
\begin{gathered}
\Rightarrow x=(2 n+1) \pi-y \quad \text { or } \quad x=2 n \pi+y, \quad \text { where } n \in \mathbb{Z} \\
\Rightarrow x=(2 n+1) \pi+(-1)^{2 n+1} y \quad \text { or } \quad x=2 n \pi+(-1)^{2 n} y, \quad \text { where } n \in \mathbb{Z}
\end{gathered}
$

Combining these two results, we get:

$
x=n \pi+(-1)^n y, \quad \text { where } n \in \mathbb{Z}
$

2. Prove that for any real numbers x and y, cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

Proof: If $\cos x=\cos y$, then

$
\begin{gathered}
\cos x-\cos y=0 \\
\Rightarrow-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=0 \\
\Rightarrow \sin \left(\frac{x+y}{2}\right)=0 \quad \text { or } \quad \sin \left(\frac{x-y}{2}\right)=0 \\
\Rightarrow \frac{x+y}{2}=n \pi \quad \text { or } \quad \frac{x-y}{2}=n \pi, \quad \text { where } n \in \mathbb{Z}
\end{gathered}
$

(Because $\sin A=0$ implies $A=n \pi$, where $n \in \mathbb{Z}$ )

$
\Rightarrow x=2 n \pi-y \quad \text { or } \quad x=2 n \pi+y, \quad \text { where } n \in \mathbb{Z}
$

Hence,

$
x=2 n \pi \pm y, \quad \text { where } n \in \mathbb{Z}
$

3. Prove that if x and y are not odd multiples of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.

Proof: If $\tan x=\tan y$, then

$
\begin{gathered}
\tan x-\tan y=0 \\
\Rightarrow \frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}=0 \\
\Rightarrow \frac{\sin x \cos y-\cos x \sin y}{\cos x \cos y}=0 \\
\Rightarrow \sin (x-y)=0
\end{gathered}
$

$\Rightarrow x-y=n \pi, \quad$ where $n \in \mathbb{Z} \quad[$ Because $\sin A=0$ implies $A=n \pi$, where $n \in \mathbb{Z}]$

$
\Rightarrow x=n \pi+y, \quad \text { where } n \in \mathbb{Z}
$

Principal Solution

The solutions of a trigonometric equation that lie in the interval [0, 2π). For example, if $2 \sin \theta=1$, then the two values of sinӨ between 0 and 2π are π/6 and 5π/6. Thus, π/6 and 5π/6 are the principal solutions of equation $2 \sin \theta=1$.

General Solution

As trigonometric functions are periodic, solutions are repeated within each period, so, trigonometric equations may have an infinite number of solutions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

We have the following trigonometric equations whose solutions are quadrantile angles.

Steps to Solve Trigonometric Equations

• Transform the given trigonometric equation into an equation with a single trigonometric ratio (sin, cos, tan).
• Change the equation with the trigonometric equation, having multiple angles, or submultiple angles into a simple angle.
• Now represent the equation as a polynomial equation, quadratic equation, or linear equation.
• Solve the trigonometric equation similar to normal equations, and find the value of the trigonometric ratio.
• The angle of the trigonometric ratio or the value of the trigonometric ratio represents the solution of the trigonometric equation.

General Solution of some Standard Equations

1. $\sin \theta=\sin \alpha$

$
\begin{aligned}
& \text { Given, } \sin \theta=\sin \alpha \Rightarrow \sin \theta-\sin \alpha=0 \\
& \Rightarrow 2 \cos \frac{\theta+\alpha}{2} \sin \frac{\theta-\alpha}{2}=0 \\
& \Rightarrow \cos \frac{\theta+\alpha}{2}=0 \quad \text { or } \quad \sin \frac{\theta-\alpha}{2}=0 \\
& \Rightarrow \frac{\theta+\alpha}{2}=(2 \mathrm{n}+1) \frac{\pi}{2} \quad \text { or } \quad \frac{\theta-\alpha}{2}=\mathrm{n} \pi, \quad \mathrm{n} \in \mathbb{I} \\
& \Rightarrow \theta=(2 \mathrm{n}+1) \pi-\alpha \quad \text { or } \theta=2 \mathrm{n} \pi+\alpha, \quad \mathrm{n} \in \mathbb{I} \\
& \Rightarrow \theta=(\text { any odd multiple of } \pi)-\alpha \space \space ..... (i)\\
& \Rightarrow \theta=(\text { any even multiple of } \pi)+\alpha \space .....(ii)\\
& \text { from (i) and (ii) } \\
& \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \mathrm{n} \in \mathbb{I}
\end{aligned}
$

2. $\cos \theta=\cos \alpha$

$\begin{aligned} & \Rightarrow \cos \alpha-\cos \theta=0 \\ & \Rightarrow 2 \sin \frac{\alpha+\theta}{2} \sin \frac{\theta-\alpha}{2}=0 \\ & \Rightarrow \sin \frac{\alpha+\theta}{2}=0 \text { or } \sin \frac{\theta-\alpha}{2}=0 \\ & \Rightarrow \frac{\alpha+\theta}{2}=\mathrm{n} \pi \text { or } \frac{\theta-\alpha}{2}=\mathrm{n} \pi, \mathrm{n} \in \mathbb{I} \\ & \Rightarrow \theta=2 \mathrm{n} \pi-\alpha \text { or } \theta=2 \mathrm{n} \pi+\alpha, \mathrm{n} \in \mathrm{Z} \\ & \Rightarrow \theta=2 \mathrm{n} \pi \pm \alpha, \mathrm{n} \in \mathbb{I}\end{aligned}$

3. $\tan \theta=\tan \alpha$

$
\begin{aligned}
& \text { Given, } \tan \theta=\tan \alpha \\
& \Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha}{\cos \alpha} \\
& \Rightarrow \sin \theta \cos \alpha-\cos \theta \sin \alpha=0 \\
& \Rightarrow \sin (\theta-\alpha)=0 \\
& \Rightarrow \theta-\alpha=\mathrm{n} \pi \\
& \Rightarrow \theta=\mathrm{n} \pi+\alpha, \text { where } \mathrm{n} \in \mathbb{I}
\end{aligned}
$

Trigonometric Equation $f^2(x)=f^2(\alpha)$, where $\mathrm{f}(\mathrm{x})$ is trigonometric function

4. $\sin ^2 \theta=\sin ^2 \alpha$

$
\begin{aligned}
& \Rightarrow \sin ^2 \theta=\sin ^2 \alpha \\
& \Rightarrow \sin (\theta+\alpha) \sin (\theta-\alpha)=0
\end{aligned}
$
$\because$ we are using the identity, $\sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B$
$
\begin{aligned}
& \Rightarrow \sin (\theta+\alpha)=0 \text { or } \sin (\theta-\alpha)=0 \\
& \Rightarrow \theta+\alpha=n \pi \text { or } \theta-\alpha=n \pi, n \in \mathbb{I} \\
& \Rightarrow \theta=n \pi \pm \alpha \in \mathbb{I}
\end{aligned}
$

Note:

The general solution of the equation $\cos ^2 \theta=\cos ^2 \alpha$ and $\tan ^2 \theta=\tan ^2 \alpha$ is also $\theta=\mathrm{n} \pi \pm \alpha \in \mathbb{I}$.

Important Points to remember while solving trigonometric equations

1. While solving a trigonometric equation, squaring the equation at any step should be avoided as much as possible. If squaring is necessary, check the solution for values that do not satisfy the original equation.
2. Never cancel terms containing unknown terms on the two sides that are in the product. It may cause the loss of a genuine solution.
3. The answer should not contain such values of angles that make any of the terms undefined or infinite.
4. Domain should not change while simplifying the equation. If it changes, necessary corrections must be made.
5. Check that the denominator is not zero at any stage while solving the equations.

Summary

Solving a triangle entails utilizing geometric and trigonometric principles to determine the lengths of sides and measures of angles based on given information. By applying the Law of Sines, the Law of Cosines, and other relevant formulas, one can systematically find all unknown elements of the triangle, ensuring accuracy through the verification of angles and side lengths.

Solved Examples Based on Solution of Trigonometric Equations

Example 1: $
\text { Let } S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}\space {\text {and }} \mathrm{b}=\displaystyle\sum_{x \in \mathrm{S}} \tan ^2\left(\frac{\mathrm{x}}{3}\right) \text {, then } \frac{1}{6}(\beta-14)^2 \text { is }
$ is equal to: [JEE MAINS 2023]

Example 2: The set of all values $\lambda$ of for which the equation $\cos ^2 2 x-2 \sin ^4 x-2 \cos ^2 x=\lambda$ has a real solution $x$, is [JEE MAINS 2023]