Question

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

• A. $n\pi +\frac{\pi }{8}$ where $nϵI$
• B. $\frac{n\pi }{2}+\frac{\pi }{8}$ where $nϵI$
• C. $(-1{)}^n\left(\begin{array}{c}\frac{n\pi }{2}+\frac{\pi }{8}\end{array}\right)$ where $nϵI$
• D. $2n\pi +co{s}^{-1}\left(\begin{array}{c}\frac{3}{2}\end{array}\right)$ where $nϵI$

Answer 1

The correct option is B $\frac{n\pi }{2}+\frac{\pi }{8}$ where $nϵI$

Given equation is
$\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x$
$(\sin x+\sin 3x)-3\sin 2x=(\cos x+\cos 3x)-3\cos 2x$
$(2\sin 2x\cos x-3\sin 2x)-(2\cos 2x\cos x-3\cos 2x)=0$
$\sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=0$
$\Rightarrow (\sin 2x-\cos 2x)(2\cos x-3)=0$
$cosx=\frac{3}{2}$
which is not possible
$\sin 2x=\cos 2x$
$\tan 2x=1$
$\tan 2x=\tan \frac{\pi }{4}$
$2x=n\pi +\frac{\pi }{4}$
$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8},n\in I$