Question

If $Si{n}^{-1}\left(6x\right)+Si{n}^{-1}\left(6\sqrt{3}x\right)=-\frac{\pi }{2}$ then the value of $x$ is

• A. $\frac{1}{12}$
• B. $-\frac{1}{12}$
• C. $-\frac{1}{4\sqrt{3}}$
• D. $\frac{1}{4\sqrt{3}}$

Answer 1

Answer: (B)

$-\frac{1}{12}$

Consider the given equation.

${\sin }^{-1}\left(6x\right)+{\sin }^{-1}\left(6\sqrt{3}x\right)=-\frac{\pi }{2}$ ……. (1)

$\frac{\pi }{2}+{\sin }^{-1}\left(6\sqrt{3}x\right)=-{\sin }^{-1}\left(6x\right)$

On taking $\sin$ both sides, we get

$\sin (\frac{\pi }{2}+{\sin }^{-1}\left(6\sqrt{3}x\right))=\sin (-{\sin }^{-1}\left(6x\right))$

We know that

$\sin (\frac{\pi }{2}+\theta )=\cos \theta$

$\sin (-\theta )=-\sin \theta$

Therefore,

$\cos \left({\sin }^{-1}\left(6\sqrt{3}x\right)\right)=-\sin \left({\sin }^{-1}\left(6x\right)\right)$

We know that

${\sin }^{-1}x={\cos }^{-1}\sqrt{(1-x^2)}$

Therefore,

$\cos \left({\sin }^{-1}\left(6\sqrt{3}x\right)\right)=-\sin \left({\sin }^{-1}\left(6x\right)\right)$

$\cos \left({\cos }^{-1}\left(\sqrt{1-{\left(6\sqrt{3}x\right)}^2}\right)\right)=-\sin \left({\sin }^{-1}\left(6x\right)\right)$

$\sqrt{1-{\left(6\sqrt{3}x\right)}^2}=-6x$

On squaring both sides, we get

$1-{\left(6\sqrt{3}x\right)}^2={(-6x)}^2$

$1-108x^2=36x^2$

$144x^2=1$

$x^2=\frac{1}{144}$

$x=\pm \frac{1}{12}$

But $x=\frac{1}{12}$ does not satisfy equation (1),

Hence, the value of $x$ is $-\frac{1}{12}$.