Question

$si{n}^{-1}6x+si{n}^{-1}6\sqrt{3}x=-\pi /2$ if x is equal to

• A. -1/12
• B. 1/6
• C. 1/12
• D. -1/6

Answer 1

The correct option is A -1/12

Given, ${\sin }^{-1}6x+{\sin }^{-1}6\sqrt{3}x=-\frac{\pi }{2}\Rightarrow {\sin }^{-1}6x=-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3}x\Rightarrow sin\left({\sin }^{-1}6x\right)=\sin (-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3}x)\Rightarrow 6x=\cos \left(si{n}^{-1}6\sqrt{3}x\right)\Rightarrow 6x=\cos \left({\cos }^{-1}\left(\sqrt{1-108x^2}\right)\right)\Rightarrow 6x=\sqrt{1-108x^2}\Rightarrow 36x^2=1-108x^2\Rightarrow 144x^2=1\Rightarrow x=\pm \frac{1}{12}$
For $x=\frac{1}{12}$
${\sin }^{-1}6x+{\sin }^{-1}6\sqrt{3}x=-\frac{\pi }{2}\Rightarrow {\sin }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{\sqrt{3}}{2}\Rightarrow \frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$
And for $x=-\frac{1}{12}$
${\sin }^{-1}6x+{\sin }^{-1}6\sqrt{3}x=-\frac{\pi }{2}\Rightarrow {\sin }^{-1}(-\frac{1}{2})+{\sin }^{-1}(-\frac{\sqrt{3}}{2})\Rightarrow -\frac{\pi }{6}-\frac{\pi }{3}=-\frac{\pi }{2}$
Hence, $x=-\frac{1}{12}$ is the only solution