Value of $x$ in inverse equation

Question

Finding value of $x$ in $\arcsin(x)+\arccos(1-x)=\arcsin(-x)$

Try: $\arcsin(x)+\arccos(1-x)=-\arcsin(x)$

$\arccos(1-x)=-2\arcsin(x)=-\arcsin(2x\sqrt{1-x^2})$

Could some help me to solve in some easy way. Thanks

Answer 1

For real cases we need $-1\le x\le1$ and $-1\le1-x\le1\iff1\ge x-1\ge-1\iff2\ge x\ge0$

$\implies0\le x\le1\ \ \ \ (1)$

As $\arcsin(-x)=-\arcsin(x)$

$$\arccos(1-x)=2\arcsin(-x)=-2\arcsin(x)$$

Again $0\le\arccos(1-x)\le\pi,0\le-2\arcsin(x)\le\pi$

$\iff0\ge\arcsin(x)\ge-\dfrac\pi2\iff0\ge x\ge-1\ \ \ \ (2)$

By $(1),(2),x=0$

Alternatively, let $\arcsin(x)=y,x=?$

$$1-\sin y=\cos(-2y)=1-2\sin^2y\iff\sin y(2\sin y-1)=0 $$

But $x=\sin y=\dfrac12$ does not satisfy the given relation