Question

The number of solutions of the equation $\sin \left(x\right)\cos \left(3x\right)=\sin \left(3x\right)\cos \left(5x\right)$ in $\left[0,\frac{\pi }{2}\right]$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C

$5$

Explanation for correct option

Given:

$$\begin{gathered} \sin \left(x\right)\cos \left(3x\right)=\sin \left(3x\right)\cos \left(5x\right) \\ \Rightarrow 2\sin \left(x\right)\cos \left(3x\right)=2\sin \left(3x\right)\cos \left(5x\right) \\ \Rightarrow \sin \left(x+3x\right)+\sin \left(x-3x\right)=\sin \left(3x+5x\right)+\sin \left(3x-5x\right) \\ \Rightarrow \sin \left(4x\right)+\sin \left(-2x\right)=\sin \left(8x\right)+\sin \left(-2x\right) \\ \Rightarrow s\mathrm{in}\left(4x\right)=\sin \left(8x\right) \\ \Rightarrow \sin \left(4x\right)=2\sin \left(4x\right)\cos \left(4x\right) \\ \Rightarrow \cos \left(4x\right)=\frac{1}{2} \\ \Rightarrow 4x={\cos }^{-1}\left(\frac{1}{2}\right) \end{gathered}$$

From the given,

$\because 0\le x\le \frac{\pi }{2}$

$\Rightarrow 0\le 4x\le 2\mathrm{\pi }$

Therefore,

$$\begin{gathered} 4x=2n\pi \pm \frac{\pi }{3} \\ \Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{12} \end{gathered}$$

Therefore, the solution are $x=0,\frac{\pi }{12},\frac{5\pi }{12},\frac{\pi }{4},\frac{\pi }{2}$.

So, the total number of solutions is $5$.

Hence, the correct option is $\left(\mathrm{C}\right)$.