Question

Solve the equation ${sin}^{-1}6x+{sin}^{-1}6\sqrt{3}x=\frac{-\pi }{2}$

Answer 1

From the given equation, we have ${\sin }^{-1}6x=\frac{-\pi }{2}-{\sin }^{-1}6\sqrt{3}x$
$\Rightarrow {\sin }^{-1}6x=-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3x}$
$\Rightarrow \sin \left({\sin }^{-1}6x\right)=\sin (-\frac{\pi }{2}-{\sin }^{-1}6\sqrt{3x})$
$\Rightarrow 6x=-\cos \left({\sin }^{-1}6\sqrt{3x}\right)$
$\Rightarrow 6x=-\sqrt{1-108x^2}$. squaring both, we get$36x^2=1-108x^2$
$\Rightarrow 144x^2=1$ $\Rightarrow x\pm \frac{1}{12}$
Note that $x=\frac{1}{12}$ is the only root of the equation as $x=\frac{1}{12}$ does not satisfy it.