Question

What is the value of x which satisfy equation $si{n}^{-1}6x+si{n}^{-1}6\sqrt{3}x=-\pi /2$

• A. $x=-1/24$
• B. $x=+1/24$
• C. $x=-1/12$
• D. $x=+1/12$

Answer 1

The correct option is C $x=-1/12$

${\sin }^{-1}6x=-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3x}$
$\Rightarrow \sin {\sin }^{-1}6x=\sin (-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3x})$
Now, let ${\sin }^{-1}6\sqrt{3x}=\theta ,\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$
$\therefore \sin \theta =6\sqrt{3x},\cos \theta =\sqrt{(1-108x^2)}\therefore \theta ={\sin }^{-1}6\sqrt{3x}={\cos }^{-1}\sqrt{(1-108x^2)}$
$6x=-\sqrt{(1-108x^2)}$, RHS is negative
Therefore x is negative
Squaring ${36x}^2=1-{108x}^2\Rightarrow x=\pm \frac{1}{12}\therefore x=-\frac{1}{12}$