Question

If $\sin 2x+\cos x=0$, then which among the following is/are true?
I. $\cos x=0$
II. $\sin x=-\frac{1}{2}$
III. $x=(2n+1)\frac{\pi }{2},n\in Z$
IV. $x=n\pi +(-1{)}^n\frac{7\pi }{6},n\in Z$

• A. 1 is true
• B. I and II are true
• C. I, II and III are true
• D. All are true

Answer 1

Answer: (D)

$\sin 2x+\cos x=0$
$\Rightarrow 2\sin x\cos x+\cos x=0$
$(\because \sin 2x=2\sin x\cos x)$
$\Rightarrow \cos x(2\sin x+1)=0$
$\Rightarrow \cos x=0\text{ or }\sin x=-\frac{1}{2}$
When $\cos x=0$,
Then, $x=(2n+1)\frac{\pi }{2}$
When $\sin x=-\frac{1}{2}$,
Then, $\sin x=-\sin \frac{\pi }{6}$
$\sin x=\sin \left(\pi +\frac{\pi }{6}\right)[\because \sin (\pi +\theta )=-\sin \theta ]$
$\sin x=\sin \frac{7\pi }{6}$
$\Rightarrow x=n\pi +(-1{)}^n\frac{7\pi }{6}(n\in Z)$