Question

How do you find all solutions of the equation $\sin (3x)\cos x-\sin x\cos (3x)=0$ ?

Answer 1

Hint: In this question, we have to find the value of x. The equation given to us consists of trigonometric functions. Therefore, we will apply the trigonometric formulas and identities to solve the same. We will first apply the trigonometric formula$\sin (a-b)=\sin a\cos b-\cos a\sin b$ in the equation and then make further calculations. After that, we will remove the sin function by taking sin inverse on both sides of the equation. Then, we will get three equations, so we will solve them separately to get the required solution to the problem.

Complete step by step answer:
According to the question, we have to solve an equation for x.
Thus, we will apply the trigonometric formula to get the solution.
The equation given to us is $\sin (3x)\cos x-\sin x\cos (3x)=0$ ------ (1)
Now, we will first apply the trigonometric formula $\sin (a-b)=\sin a\cos b-\cos a\sin b$ in equation (1), we get
$\Rightarrow \sin (3x-x)=0$
Now, on further solving, we get
$\Rightarrow \sin (2x)=0$
Now, we will take ${{\sin }^{-1}}$ on both sides in the above equation, we get
$\Rightarrow {{\sin }^{-1}}\left( \sin (2x) \right)={{\sin }^{-1}}\left( 0 \right)$
Now, we know that ${{\sin }^{-1}}\left( \sin \theta \right)=\theta $ , so we will apply this formula on the left-hand side of the above equation, we get
$\Rightarrow 2x={{\sin }^{-1}}\left( 0 \right)$
Also, ${{\sin }^{-1}}\left( 0 \right)$ has three values, therefore, we get
$\Rightarrow 2x=0,\pi ,2\pi $
Therefore, we get three separate equations, that is
$\Rightarrow 2x=0$ --------- (2)
$\Rightarrow 2x=\pi $ -------- (3)
$\Rightarrow 2x=2\pi $ ------------ (4)
Now, we will solve equation (2), which is
$\Rightarrow 2x=0$
Now, divide both sides of the above equation by 2, we get
$\Rightarrow \dfrac{2}{2}x=\dfrac{0}{2}$
On further solving, we get
$\Rightarrow x=0$
Now, we will solve equation (3), which is
$\Rightarrow 2x=\pi $
So, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}x=\dfrac{\pi }{2}$
On further solving, we get
$\Rightarrow x=\dfrac{\pi }{2}$
Now, we will solve equation (4), which is
$\Rightarrow 2x=2\pi $
So, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}x=\dfrac{2\pi }{2}$
On further solving, we get
$\Rightarrow x=\pi $

Therefore, for the equation $\sin (3x)\cos x-\sin x\cos (3x)=0$ , the value of x are $0\text{, }\dfrac{\pi }{2}\text{, }and\text{ }\pi $ .

Note: In this problem, keep in mind which trigonometric formula you are using and mention them to avoid errors and mathematical calculations. While solving ${{\sin }^{-1}}\left( 0 \right)$ , do not forget that you will get three values separately and not one value. Since sin function is a periodic function that repeats after every interval $\text{0 to 2 }\!\!\pi\!\!\text{ }$ , that is why we only write three values separately.