Question

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

Answer 1

$si{n}^{-1}x-co{s}^{-1}x=\frac{\pi }{6}$

As $co{s}^{-1}x+si{n}^{-1}x=\frac{\pi }{2}$ ; $x\in [-1,1]$

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}x$

$\Rightarrow si{n}^{-1}x-(\frac{\pi }{2}-si{n}^{-1}x)=\frac{\pi }{6}$

$\Rightarrow si{n}^{-1}x-\frac{\pi }{2}+si{n}^{-1}x=\frac{\pi }{6}$

$\Rightarrow 2si{n}^{-1}x=\frac{\pi }{6}+\frac{\pi }{2}=\frac{2\pi }{3}$

$\Rightarrow si{n}^{-1}x=\frac{\pi }{3}$

$\Rightarrow x=sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}$