<s>[INST] 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
 [/INST] First of all we need $-1\le x\le1$ for real $\arcsin(x)$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
the equality will be valid
if $$x^2+\dfrac12\le1\iff x^2\le1-\dfrac12\iff?\le x\le?$$
Or if $x^2+\dfrac12>1$ and $(\sqrt2x<0\iff x<0)$
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