Question

For $n\in Z$, the general solution of the trigonometric equation $\sin x-\sqrt{3}\cos x+4\sin 2x-4\sqrt{3}\cos 2x+\sin 3x-\sqrt{3}\cos 3x=0$ is

• A. $\frac{n\pi }{2}+\frac{\pi }{8}$
• B. $\frac{n\pi }{2}+\frac{\pi }{6}$
• C. $\frac{n\pi }{2}\pm \frac{\pi }{6}$
• D. $2n\pi \pm \frac{\pi }{6}$

Answer 1

Answer: (B)

We have,
$\sin x-\sqrt{3}\cos x+4\sin 2x-4\sqrt{3}\cos 2x+\sin 3x-\sqrt{3}\cos 3x=0$
$\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x+\frac{4}{2}\sin 2x-\frac{4\sqrt{3}}{2}\cos 2x$
$+\frac{1}{2}\sin 3x-\frac{\sqrt{3}}{2}\cos 3x-0$
$\sin \left(x-\frac{\pi }{3}\right)+4\sin \left(2x-\frac{\pi }{3}\right)+\sin \left(3x-\frac{\pi }{3}\right)=0$
$\sin \left(x-\frac{\pi }{3}\right)+\sin \left(3x-\frac{\pi }{3}\right)+4\sin \left(2x-\frac{\pi }{3}\right)=0$
$2\sin \left(2x-\frac{\pi }{3}\right)\cos x+4\sin \left(2x-\frac{\pi }{3}\right)=0$
$2\sin \left(2x-\frac{\pi }{3}\right)(\cos x+2)=0$
$\sin \left(2x-\frac{\pi }{3}\right)=0,\cos x+2\ne 0$
$\Rightarrow 2x-\frac{\pi }{3}=n\pi$
$\Rightarrow 2x=n\pi +\frac{\pi }{3}$
$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{6}$