Question

The general solution of $\sin 3x+\sin x-3\sin 2x=\cos 3x+\cos x-3\cos 2x$ is

• A. $\left(n\frac{\pi }{2}\right)+\left(\frac{\pi }{8}\right)$ for $n$ integer
• B. $\left(n\frac{\pi }{2}\right)-\left(\frac{\pi }{8}\right)$ for $n$ integer
• C. $n\pi +\left(\frac{\pi }{6}\right)$ for $n$ integer
• D. $n\pi -\left(\frac{\pi }{6}\right)$ for $n$ integer

Answer 1

The correct option is A

$\left(n\frac{\pi }{2}\right)+\left(\frac{\pi }{8}\right)$ for $n$ integer

Explanation for the correct answer:

Find the general solution

Given: $\sin 3x+\sin x-3\sin 2x=\cos 3x+\cos x-3\cos 2x$$$\begin{gathered} \Rightarrow (s\mathrm{in}x+\sin 3x)-3\sin 2x=(\cos x+\cos 3x)-3\cos 2x \\ \Rightarrow 2sin2x\cos x-3\sin 2x=2\cos 2x\cos x-3cos2x\left[\because \sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\&\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)\right] \\ \Rightarrow \sin 2x(2\cos x-3)=\cos 2x(2\mathrm{co}sx-3) \\ \Rightarrow \sin 2x=\cos 2x \\ \Rightarrow \frac{\sin 2x}{\cos 2x}=1 \\ \Rightarrow \tan 2x=1\left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \right] \\ \Rightarrow \tan 2x=\tan \frac{\mathrm{\pi }}{4}\left[\because \tan \frac{\mathrm{\pi }}{4}=1\right] \\ \Rightarrow 2x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=\mathrm{n}\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{8},n\in \mathrm{\mathbb{Z}} \end{gathered}$$

Hence, option (A) $\left(n\frac{\pi }{2}\right)+\left(\frac{\pi }{8}\right)$, is the correct answer.