Question

Solve \(sin^{-1}(1-x)-2sin^{-1}x=\frac {\pi}{2}\), then x is equal to

• A. \(0, \frac 12\)
• B. \(1,\frac 12\)
• C. \(0\)
• D. \(\frac 12\)

Answer 1

The Correct Option is C

sin^-1(1-x) - 2sin^-1x = \(\frac {\pi}{2}\) 
\(\implies\)-2 sin^-1x = \(\frac {\pi}{2}\) - sin^-1(1-x) 
\(\implies\)-2 sin^-1x = cos^-1(1-x) .………. (1) 
Let sin^-1x = θ\(\implies\)sinθ = x\(\implies\)cosθ = \(\sqrt {1-x^2}\). 
θ = cos^-1\((\sqrt {1-x^2})\) 
Therefore, from equation (1), we have;
-2 cos^-1\((\sqrt {1-x^2})\) = cos^-1(1-x) 
Put x = sin y. Then, we have:
-2 cos^-1\((\sqrt {1-sin^2 y})\) = cos^-1(1-sin y) 
\(\implies\)-2 cos-¹(cos y) = cos^-1(1-sin y) 
\(\implies\)-2y = cos^-1(1-sin y) 
\(\implies\)1-siny = cos(-2y) = cos 2y 
\(\implies\)1-sin y = 1-2sin²y 
\(\implies\)2sin²y - sin y = 0 
\(\implies\)sin y (2 sin y-1) = 0 siny = 0 or \(\frac 12\) 
therefore x = 0 or \(\frac 12\) 
But, when x = \(\frac 12\) , it can be observed that: 
L.H.S = sin^-1(1-\(\frac 12\)) - 2 sin^-1 \(\frac 12\) 
        = sin^-1\((\frac 12)\)-2 sin^-1\((\frac 12)\) 
        = - sin^-1\(\frac 12\) 
        = \(-\frac {\pi}{6}\) ≠ \(-\frac {\pi}{2}\)
        ≠ R.H.S 
so x = \(\frac 12\) is not the solution of the given equation. Thus, x = 0. 

Hence, the correct answer is (C): \(0\)