A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{8}\]
D) \[\frac{\pi }{4}\]
E) \[\frac{\pi }{12}\]
Correct Answer: E
Solution :
We have \[\sin x\cos x=\frac{1}{4}\] \[\Rightarrow \] \[\sin 2x=\frac{1}{2}\] \[\Rightarrow \] \[\sin 2x=\sin \frac{\pi }{6}\] \[\Rightarrow \] \[2x=nx+{{(-1)}^{n}}\frac{\pi }{6},n\in Z\] \[\Rightarrow \] \[x=\frac{n\pi }{2}+{{(-1)}^{n}}\frac{\pi }{12},n\in z\] \[\Rightarrow \] \[x=\frac{\pi }{12}\in \left( 0,\frac{\pi }{12} \right)for\,n=0\]
You need to login to perform this action.
You will be redirected in
3 sec
