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