Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}}  - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}}  + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}}  + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}}  + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here

Substitute $x = \sin\theta$ and let's restrict our attention to $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ since that covers the whole range of $\sin\theta$:
$$\sin^{-1}\left(\frac{\sin\theta+\cos\theta}{\sqrt{2}}\right) = \frac{\pi}{4}+\sin^{-1}(\sin\theta)$$
$$\implies \sin^{-1}\left(\sin\left(\theta+\frac{\pi}{4}\right)\right) = \frac{\pi}{4}+\theta$$
However, the equality is only true when $$\left|\theta+\frac{\pi}{4}\right|\leq \frac{\pi}{2} \implies -\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}$$
Taking the intersection of this set with our original domain, we have that
$$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{4} \implies -1 \leq x \leq \frac{1}{\sqrt{2}}$$
