Question

If ${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6}$, then solve for $x$.

Answer 1

We have

${\sin }^{-1}x-{\cos }^{-1}x=\pi /6$

$\Rightarrow {\sin }^{-1}x+{\cos }^{-1}x-2\cdot {\cos }^{-1}x=\pi /6$

$\Rightarrow \pi /2-2\cdot {\cos }^{-1}x=\pi /6$

$\Rightarrow -2\cdot {\cos }^{-1}x=\pi /6-\pi /2$

$\Rightarrow -2\cdot {\cos }^{-1}x=\frac{\pi -3\pi }{6}$

$\Rightarrow -2\cdot {\cos }^{-1}x=\frac{\pi }{6}-\frac{3\pi }{6}$

$\Rightarrow 2\cdot {\cos }^{-1}x=2\pi /6$

$\Rightarrow {\cos }^{-1}x=\frac{2\pi }{6.2}$

$\Rightarrow {\cos }^{-1}x=\pi /6$

$\Rightarrow x=\cos \pi /6$

$\Rightarrow x=\sqrt{3}/2$.