Question

2 sin−1(x) + cos−1(x) = π
The hint is to take the cosine of each side

Answer 1

First take $y=\sin^{-1}x$ and $z=\cos^{-1}x$. Then

$\cos(2\sin^{-1}x+\cos^{-1}x)=\cos(2y+z)=\cos2y\cos z-\sin2y\sin z$
$=(\cos^2y-\sin^2y)\cos z-2\sin y\cos y\sin z$

Now

$\sin y=\sin(\sin^{-1}x)=x$
$\cos y=\cos(\sin^{-1}x)=\sqrt{1-x^2}$

both provided that $-\dfrac\pi2\le y\le\dfrac\pi2$, and

$\sin z=\sin(\cos^{-1}x)=\sqrt{1-x^2}$
$\cos z=\cos(\cos^{-1}x)=x$

both provided that $0\le z\le\pi$.

Then

$\cos(2\sin^{-1}x+\cos^{-1}x)=(1-2x^2)x-2x(1-x^2)$
$\implies -x=\cos\pi\implies -x=-1\implies x=1$