Question

The possible values of $\theta \in (0,\pi )$ such that $\sin \left(\theta \right)+\sin \left(4\theta \right)+\sin \left(7\theta \right)=0$ are:

• A. $\frac{2\pi }{9},\frac{\pi }{4},\frac{4\pi }{9},\frac{\pi }{2},\frac{3\pi }{4},\frac{8\pi }{9}$
• B. $\frac{\pi }{4},\frac{5\pi }{12},\frac{\pi }{2},\frac{2\pi }{3},\frac{3\pi }{4},\frac{8\pi }{9}$
• C. $\frac{2\pi }{9},\frac{\pi }{4},\frac{\pi }{2},\frac{2\pi }{3},\frac{3\pi }{4},\frac{35\pi }{36}$
• D. $\frac{2\pi }{9},\frac{\pi }{4},\frac{\pi }{2},\frac{2\pi }{3},\frac{3\pi }{4},\frac{8\pi }{9}$

Answer 1

The correct option is A $\frac{2\pi }{9},\frac{\pi }{4},\frac{4\pi }{9},\frac{\pi }{2},\frac{3\pi }{4},\frac{8\pi }{9}$

We have $sin\theta +sin4\theta +sin7\theta =0$

$\Rightarrow [sin7\theta +sin\theta ]+sin4\theta =0$

$\Rightarrow 2sin\left(\frac{7\theta +\theta }{2}\right)cos\left(\frac{7\theta -\theta }{2}\right)+sin4\theta =0[\because sinA+sinB=2sin(\frac{A+B}{2}\left)cos\right(\frac{A-B}{2}\left)\right]$

$\Rightarrow 2sin4\theta cos3\theta +sin4\theta =0$

$\Rightarrow sin4\theta [2cos3\theta +1]=0$

$\Rightarrow sin4\theta =0$ or $2cos3\theta +1=0$

When $sin4\theta =0$, then

$4\theta =\pi ,2\pi ,3\pi ,4\pi ,...$

$\Rightarrow \theta =\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\pi ,...$

When $2cos3\theta +1=0$

$\Rightarrow cos3\theta =-\frac{1}{2}$

$\Rightarrow 3\theta =\pi -\frac{\pi }{3},\pi +\frac{\pi }{3},3\pi -\frac{\pi }{3},3\pi +\frac{\pi }{3},...[\because cos\frac{\pi }{3}=\frac{1}{2}]$

$\Rightarrow 3\theta =\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{11\pi }{3},...$

$\Rightarrow \theta =\frac{2\pi }{9},\frac{4\pi }{9},\frac{8\pi }{9},\frac{11\pi }{9},...$

As $\theta \in (0,\pi )$

$\theta =\frac{2\pi }{9},\frac{\pi }{4},\frac{4\pi }{9},\frac{\pi }{2},\frac{3\pi }{4},\frac{8\pi }{9}$

So, $\text{A}$ is the correct option.