Question

For $n \in Z$, the general solution of the trigonometric equation $\sin x-\sqrt{3} \cos x +4 \sin 2 x-4 \sqrt{3} \cos 2 x+\sin 3 x-\sqrt{3} \cos 3 x=0$ 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. $2 n \pi \pm \frac{\pi}{6}$

Answer 1

Answer: (B)

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