Question

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

• A. $n\pi +\frac{\pi }{8}$
• B. $\frac{n\pi }{2}+\frac{\pi }{8}$
• C. $(-1{)}^n\frac{n\pi }{2}+\frac{\pi }{8}$
• D. $2n\pi +{\cos }^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (B)

$\sin x-3\sin 2x+\sin 3x$
$=\cos x-3\cos 2x+\cos 3x$
$\Rightarrow (\sin 3x+\sin x)-3\sin 2x$
$=(\cos 3x+\cos x)-3\cos 2x$
$\Rightarrow 2\sin 2x\cos x-3\sin 2x$
$=2\cos 2x\cos x-3\cos 2x$
$\Rightarrow \sin 2x(2\cos x-3)=\cos 2x(2\cos x-3)$
$\Rightarrow \sin 2x=\cos 2x[\because \cos x-3\ne 0]$
$\Rightarrow \tan 2x=1=\tan \frac{\pi }{4}$
$\Rightarrow 2x=n\pi +\frac{\pi }{4}\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8}$