Question

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