<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] The other answers pointed out some very creative ways of solving your problem, but I just wanted to complete your algebraic manipulation for you.
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}}  + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt 
{\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
${x\sqrt {1 - {x^2}}  + 1 - {x^2} - x\sqrt 
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = 1$
${x\sqrt {1 - {x^2}} - x\sqrt 
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = x^2$
${\sqrt {1 - {x^2}} - \sqrt 
{\left( {1 - 2x\sqrt {1 - {x^2}}} \right)} } = x$
(Since we divide both sides by $x$, you can verify by hand that $x = 0$ is a solution)
${\sqrt {1 - {x^2}} - \sqrt 
{\left( {x - \sqrt{1 - {x^2}}} \right)^2} } = x$
${\sqrt {1 - {x^2}} - x + \sqrt{1 - {x^2}}} = x$
$2\sqrt {1 - {x^2}} = 2x$
$\sqrt {1 - {x^2}} = x$
$1 - {x^2} = x^2$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$
Thus, we are left with three solutions: $\{ -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\}$
Only option choice (C) contains all three solutions found.
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