Concept:

\({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\) and \({\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x\)

Calculation:

\({\sin ^{ - 1}}\left( {1 - x} \right) + 2{\sin ^{ - 1}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right) = \frac{\pi }{2}\)

\({\sin ^{ - 1}}\left( {1 - x} \right) + 2{\sin ^{ - 1}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right) = {\sin ^{^{ - 1}}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right)\)

\({\sin ^{ - 1}}\left( {1 - x} \right) + {\sin ^{ - 1}}2x = 0\)

\({\sin ^{ - 1}}\left( {1 - x} \right) = - {\sin ^{ - 1}}\left( {2x} \right)\)

\({\sin ^{ - 1}}\left( {1 - x} \right) = {\sin ^{ - 1}}\left( { - 2x} \right)\)

\(1 - x = - 2x\)

\(x = - 1\)

But as we can see that, by substituting x = - 1 in the given equation we get, sin^-1(2) + sin^-1(-2) = 0, but we know that domain of sin^-1 x is [-1, 1]

So, x = - 1 is not possible solution of the given equation.

Hence, there is no value of x which satisfies the given equation