Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$, then general
value of $\theta$ is
Options:
value of $\theta$ is
Solution:
2066 Upvotes
Verified Answer
The correct answer is:
$\frac{n \pi}{4}, n \pi \pm \frac{\pi}{3}$
We have, $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$
$\Rightarrow \quad \sin 6 \theta+\sin 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad 2 \sin 4 \theta \cdot \cos 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad \sin 4 \theta(2 \cos 2 \theta+1)=0$
$\Rightarrow \quad \sin 4 \theta=0$ or $2 \cos 2 \theta+1=0$
$\Rightarrow \quad 4 \theta=n \pi$ or $\cos 2 \theta=\frac{-1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $2 \theta=2 n \pi \pm \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\theta=n \pi \pm \frac{\pi}{3},$ where $n \in z$
$\Rightarrow \quad \sin 6 \theta+\sin 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad 2 \sin 4 \theta \cdot \cos 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad \sin 4 \theta(2 \cos 2 \theta+1)=0$
$\Rightarrow \quad \sin 4 \theta=0$ or $2 \cos 2 \theta+1=0$
$\Rightarrow \quad 4 \theta=n \pi$ or $\cos 2 \theta=\frac{-1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $2 \theta=2 n \pi \pm \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\theta=n \pi \pm \frac{\pi}{3},$ where $n \in z$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.