Question

Evaluate: ${\sin }^{-1}6x+{\sin }^{-1}6\sqrt{3}x=-\frac{\pi }{2}$

Answer 1

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

$\frac{\pi }{2}-{\text{cos}}^{-1}6x+{\text{sin}}^{-1}6\sqrt{3}x=-\frac{\pi }{2}$

$⟹\pi +{\text{sin}}^{-1}6\sqrt{3}x={\text{cos}}^{-1}6x$

$⟹\cos (\pi +{\text{sin}}^{-1}6\sqrt{3}x)=6x$

$⟹-\sqrt{1-108x^2}=6x⟹x<0$

$1-108x^2=36x^2⟹x^2=\frac{1}{144}⟹x=-\frac{1}{12}$

$\because x=\frac{1}{12}$ does not satisfy the equation.