If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$, then general value of $\theta$ is
If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$, then general
value of $\theta$ is
- $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{3}$
- $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{6}$
- $\frac{n \pi}{4}, 2 n \pi \pm \frac{\pi}{3}$
- $\frac{n \pi}{4}, 2 n \pi \pm \frac{\pi}{6}$
Solution
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 $n \in z$
Asked in: TEST SERIES MHT-CET Full Test 6
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