Question

The general solution of $\sin x-3sin2x+sin3x=cosx-3cos2x+cos3x$ is-

• A. $n\pi +\pi /8$
• B. $n\pi /2+\pi /8$
• C. $(-1{)}^n(n\pi /2+\pi /8)$
• D. $2n\pi +co{s}^{-1}\left(3/2\right)$

Answer 1

The correct option is B $n\pi /2+\pi /8$

$\sin x-3\sin 2x+sin3x=\cos x-3\cos 2x+\cos 3x$
$\Rightarrow 2\sin 2x\cos x-3sin2x=2\cos 2x\cos x-3\cos 2x$
$\Rightarrow (\sin 2x-\cos 2x)(2\cos x-3)=0$
$\Rightarrow \cos x=\frac{3}{2}$ (not possible) $\tan 2x=1$
$\Rightarrow \tan 2x=\tan \frac{\pi }{4}$
So, the general solution is
$x=\frac{n\pi }{2}+\frac{\pi }{8}$