# Solve the equation: $\sin^{-1}(6x) + \sin^{-1}(6\sqrt 3 x) =-\large \frac{\pi}{2}$

(A) $x=-\large\frac{1}{2}$ (B) $x=\large\frac{1}{12}$ (C) $x=-\large\frac{1}{6}$ (D) $x=-\large\frac{1}{12}$

Toolbox:
• $sin(-\theta)=-sin\theta$
• $sin(\large\frac{\pi}{2}$$+\theta)=cos\theta • sin^{-1}x=cos^{-1}\sqrt{1-x^2} • sin^{-1}(sin\theta)=\theta • cos^{-1}(cos\theta)=\theta Given eqn. is written as -sin^{-1}6x=\large\frac{\pi}{2} +sin^{-1}(6\sqrt 3 x) Taking sin on both the sides we get sin(-sin^{-1}6x)=sin(\large\frac{\pi}{2}$$+sin^{-1}6\sqrt 3 x)$
We know that $sin(-\theta)=-sin\theta$
$\Rightarrow\:-sin(sin^{-1}6x)=cos(sin^{-1}6\sqrt 3 x)$
But we know that $sin^{-1}x=cos^{-1}\sqrt{1-x^2}$
$\Rightarrow \:sin^{-1}6\sqrt 3 x=cos^{-1}\sqrt{1-(6\sqrt 3 x)^2}=cos^{-1}\sqrt{1-108x^2}$
and we also know that $sin^{-1}(sin\theta)=\theta$
$\Rightarrow\:-6x=cos(cos^{-1}\sqrt{1-108x^2}$
$\Rightarrow\: -6x=\sqrt{1-108x^2}$
Squaring on both the sides we get
$36x^2=1-108x^2$
$\Rightarrow\: 144x^2=1$
$\Rightarrow \:x^2=\large\frac{1}{144}$
$\Rightarrow x=\pm\large\frac{1}{12}$
But $x=\large\frac{1}{12}$ does not satisfy the eqn.
$\therefore\: x=-\large\frac{1}{12}$