Answer
$2sin^{-1}~x = cos^{-1}(1-2x^2)~~~~$ $x \geq 0$
Work Step by Step
Let $f(x) = 2sin^{-1}~x-cos^{-1}(1-2x^2)~~~~$ $x \geq 0$
We can find $f'(x)$:
$f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{-(-4x)}{\sqrt{1-(1-2x^2)^2}}$
$f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{\sqrt{1-(1-4x^2+4x^4)}}$
$f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{\sqrt{4x^2-4x^4}}$
$f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{2x\sqrt{1-x^2}}$
$f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{2}{\sqrt{1-x^2}}$
$f'(x) = 0$
Therefore, $f(x) = C$, where $C$ is some constant.
Consider $f(0)$:
$f(0) = 2sin^{-1}~0-cos^{-1}(1-2(0)^2)$
$f(0) = 2(0)-cos^{-1}(1)$
$f(0) = 0-0$
$f(0) = 0$
Thus $C = 0$
Then:
$f(x) = 2sin^{-1}~x-cos^{-1}(1-2x^2) = 0$
$2sin^{-1}~x = cos^{-1}(1-2x^2)~~~~$ $x \geq 0$
