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 $1-2\sqrt{2}{\sin }^{3/2}\theta =0,{\sin }^{3/2}\theta =\frac{1}{2\sqrt{2}}$
$\Rightarrow (\sin \theta {)}^{3/2}={\left(\frac{1}{2}\right)}^{\frac{3}{2}}$
$\Rightarrow \sin \theta =\frac{1}{2}=\sin \frac{\pi }{6}$
$\Rightarrow \theta =n\pi +(-1{)}^n\frac{\pi }{6},n\in I$.
$\therefore \theta =n\pi ,n\pi +(-1{)}^n\frac{\pi }{6},n\in I$.