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