Solving the equation $\arccos(x)-\arcsin(x)=\arccos(\sqrt{3}x)$

Question

I am attempting to solve the equation $\arccos(x)-\arcsin(x)=\arccos(\sqrt{3}x)$. Trying to use range of left side expression, since it is always decreasing, the range of the $\arccos(x)-\arcsin(x)$ is $\left[\pi/2-2\arcsin(1/\sqrt{3}),\pi/2+2\arcsin(1/\sqrt{3})\right]$. But that expression does not seem useful for comparison with the range of the expression $\arccos(\sqrt{3}x)$. I need to solve this without using the formula of $\arccos(x)-\arccos(y)$.

Any hints are appreciated. Thanks.

Answer 1

Hint: apply cos to both sides: $x\sqrt{1-x^2} + \sqrt{1-x^2}x = x\sqrt{3}$.

Answer 2

Hint

WLOG $\arcsin x=y,\arccos x=\dfrac\pi2-y$ and $-\dfrac\pi2\le\dfrac\pi2$

So we have

$$\dfrac\pi2-2y=\arccos(\sqrt3\sin y)\implies|\sin y|\le\dfrac1{\sqrt3}$$

Applying cosine in both sides

$$\sqrt3\sin y=\cos(\pi/2-2y)=\cdots=2\sin y\cos y$$

Can you take it from here?