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)

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