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

Answer: (C)

$0$

Answer 2

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): $0$