General Solution of Trigonometric Equations - Practice Questions & MCQ

Trigonometric Equations

Trigonometric equations are, as the name implies, equations that involve trigonometric functions.
Solution of Trigonometric Equation
The value of an unknown angle which 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 $(360+30)^0,(720+30)^0$, $(-360+30)^0$ and so on, as its solutions.

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

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

Some Important General Solutions of Equations
Equations Solution 
$\sin \theta=0$ & $\theta=n \pi, \quad n \in \mathbb{I}$
$\cos \theta=0$ & $\theta=(2 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}$ 
$\tan \theta=0$ & $\theta=n \pi, \quad n \in \mathbb{I}$ 
$\sin \theta=1$ & $\theta=(4 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}$ 
$\cos \theta=1$ & $\theta=2 n \pi, \quad n \in \mathbb{I}$ 
$\sin \theta=-1$ & $\theta=(4 n-1) \frac{\pi}{2}, \quad n \in \mathbb{I}$ 
$\cos \theta=-1$ & $\theta=(2 n+1) \pi, \quad n \in \mathbb{I}$ 
$\cot \theta=0$ & $\theta=(2 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}$

General Solution of some Standard Equations (Part 1)
1. $\sin \theta=\sin \alpha$

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

$
\theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \quad \mathrm{n} \in \mathbb{I}
$

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

$\begin{aligned} & \text { 3. } \tan \theta=\tan \alpha \\ & \text { Given, } \tan \theta=\tan \alpha \\ & \Rightarrow \quad \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha}{\cos \alpha} \\ & \Rightarrow \quad \sin \theta \cos \alpha-\cos \theta \sin \alpha=0 \\ & \Rightarrow \quad \sin (\theta-\alpha)=0 \\ & \Rightarrow \quad \theta-\alpha=\mathrm{n} \pi \\ & \Rightarrow \quad \theta=\mathrm{n} \pi+\alpha \text {, where } \mathrm{n} \in \mathbb{I}\end{aligned}$

General Solution of some Standard Equations (Part 2)

$
\begin{aligned}
& \text { 4. } \sin ^2 \boldsymbol{\theta}=\sin ^2 \mathbf{a} \\
& \Rightarrow \quad \sin ^2 \theta=\sin ^2 \dot{\alpha} \\
& \Rightarrow \quad \sin (\theta+\alpha) \sin (\theta-\alpha)=0 \\
& \because \text { we are using the identity, } \sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})=\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B} \\
& \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

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. 

Never cancel terms containing unknown terms on the two sides which are in the product. It may cause the loss of a genuine solution.

The answer should not contain such values of angles which make any of the terms undefined or infinite. 

Domain should not change while simplifying the equation. If it changes, necessary corrections must be made.

Check that the denominator is not zero at any stage while solving the equations.

Example:

Solve  $\sin x+\cos x=1$

Solution:

Given equation is, $\sin x+\cos x=1$
If we square both sides,

$
\begin{aligned}
& (\sin \mathrm{x}+\cos \mathrm{x})^2=1^2 \\
& \Rightarrow \quad \sin ^2 \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}+\cos ^2 \mathrm{x}=1 \\
& \Rightarrow \quad \sin 2 \mathrm{x}=0 \\
& \Rightarrow \quad 2 \mathrm{x}=\mathrm{n} \pi, \mathrm{n} \in \mathbb{I} \\
& \Rightarrow \quad \mathrm{x}=(\mathrm{n} \pi) / 2, \mathrm{n} \in \mathbb{I}
\end{aligned}
$
But for $n=2,6,10, \ldots$
$\sin x+\cos +=-1$ which contradicts the given equation.
Also for $n=3,7,11, \ldots$

$
\sin x+\cos x=-1
$
Hence only $n=0,4,8,12, \ldots$. and $n=1,5,9, \ldots$ satisfy the equation
Hence, the solution is $x=4 n \pi / 2=2 n \pi$ or $x=(4 n+1) \frac{\pi}{2}$