Question
Solving $$2\,{\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right),{\text{ we get}}$$
A.
$$x \in \left[ {\frac{{\sqrt 2 }}{2},1} \right]$$
B.
$$x = 3$$
C.
$$x \in \left[ {3,4} \right]$$
D.
$$x = 0$$
Answer :
$$x \in \left[ {\frac{{\sqrt 2 }}{2},1} \right]$$
Solution :
Given equation is an identity, except the range should be same for both sides
The range of $${\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right){\text{ is }}\left[ { - \frac{\pi }{2},\,\frac{\pi }{2}} \right]$$
And the range of $$2{\cos ^{ - 1}}x{\text{ is }}\left[ {0,\,2\pi } \right]$$
So the common range is $$\left[ {0,\,\frac{\pi }{2}} \right]$$
$$\eqalign{
& \Rightarrow 0 \leqslant 2{\cos ^{ - 1}}x \leqslant \frac{\pi }{2} \cr
& \Rightarrow 0 \leqslant {\cos ^{ - 1}}x \leqslant \frac{\pi }{4} \cr
& \Rightarrow \frac{1}{{\sqrt 2 }} \leqslant x \leqslant 1, \cr} $$
Since $${\cos^{ - 1}}$$ is decreasing function so, inequality will get reversed.