Question

The general value of $\theta$ satisfying the equation $2 \cos 2 \theta+\sqrt{2 \sin \theta}=2$ is -

• A. $n \pi, n \in I$
• B. $n \pi+(-1)^n \frac{\pi}{3}, n \in I$
• C. $n \pi+(-1)^n \frac{\pi}{6}, n \in I$
• D. $n \pi+(-1)^n \frac{\pi}{4}, n \in I$

Answer 1

Answer: (A), (C)

We have, $2 \cos 2 \theta+\sqrt{2 \sin \theta}=2$
$\Rightarrow \sqrt{2 \sin \theta}=2(1-\cos 2 \theta) $
$\Rightarrow \sqrt{2 \sin \theta}=4 \sin ^2 \theta $
$ \Rightarrow \sqrt{2 \sin \theta}-4 \sin ^2 \theta=0 $
$\Rightarrow \sqrt{2 \sin \theta}\left[1-2 \sqrt{2} \sin ^{3 / 2} \theta\right]=0 $
$\text { If } \sqrt{2 \sin \theta}=0, \sin \theta=0 [\because \sqrt{z}=0 \Rightarrow z=0] $
$\Rightarrow \theta=n \pi, n \in I .$
If $ 1-2 \sqrt{2} \sin ^{3 / 2} \theta=0, \sin ^{3 / 2} \theta=\frac{1}{2 \sqrt{2}} $
$ \Rightarrow(\sin \theta)^{3 / 2}=\left(\frac{1}{2}\right)^{\frac{3}{2}}$
$\Rightarrow \sin \theta=\frac{1}{2}=\sin \frac{\pi}{6} $
$\Rightarrow \theta=n \pi+(-1)^n \frac{\pi}{6}, n \in I$.
$\therefore \theta=n \pi, n \pi+(-1)^n \frac{\pi}{6}, n \in I$.