Question

The general solution of $\sin 2x+\sin 4x+\sin 6x=0$ is/are

• A. $n\pi \pm \frac{\pi }{6},n\in Z$
• B. $\frac{n\pi }{4},n\in Z$
• C. $2n\pi \pm \frac{2\pi }{3},n\in Z$
• D. $n\pi \pm \frac{\pi }{3},n\in Z$

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{n\pi }{4},n\in Z$
D $n\pi \pm \frac{\pi }{3},n\in Z$

Given: $\sin 2x+\sin 4x+\sin 6x=0$
$\Rightarrow (\sin 6x+\sin 2x)+\sin 4x=0\Rightarrow 2\sin 4x\cos 2x+\sin 4x=0\Rightarrow \sin 4x(2\cos 2x+1)=0\Rightarrow \sin 4x=0\text{or}2\cos 2x+1=0\Rightarrow \sin 4x=0\text{or}\cos 2x=-\frac{1}{2}\Rightarrow \sin 4x=0\text{or}\cos 2x=\cos \frac{2\pi }{3}\Rightarrow 4x=n\pi \text{or}2x=(2n\pi \pm \frac{2\pi }{3}),n\in Z$

Hence, the general solution is
$x=\frac{n\pi }{4}\text{or}x=(n\pi \pm \frac{\pi }{3}),n\in Z$