Question

The solution of the equation $\sin^{-1} 6x + \sin^{-1} 6 \sqrt{3} x = - \frac{\pi}{2}$ is

• A. $ - \frac{1}{12}$
• B. 12
• C. -2
• D. $ - \frac{1}{6}$

Answer 1

Answer: (A)

$\sin^{-1} 6x = - \frac{\pi}{2} - \sin^{-1} 6 \sqrt{3} x$
$ 6x = \sin\left( - \frac{\pi}{2} - \sin^{-1} 6 \sqrt{3} x \right) $ .....(1)
Now, let $\sin^{-1} 6\sqrt{3 } x = \theta, - \frac{\pi}{2} \le\theta \le \frac{\pi}{2}$
$ \therefore \sin\theta = 6 \sqrt{3} x , \cos \theta = \sqrt{1-108 x^{2}}$
$ \therefore \theta = \sin^{-1} 6 \sqrt{3} x = \cos^{-1} \sqrt{1 - 108 x^{2}} $ .....(2)
$\therefore $ R.H.S. is negative [from(2)]
$\therefore $ x is negative.
Squaring, $36 x^{2} =1 -108 x^{2}$
$ \Rightarrow x = \pm \frac{1}{12} , \therefore x = - \frac{1}{12} . $