Trigonometric equation problem

Question

This is the following equation:

$$\arccos x= \arctan x$$

Could someone give me at least a tip how to begin with?

Answer 1

Both functions are monotonic. So there is only one root for it. We can try to find it as a root for reciprocal functions. We can solve $$\cos x = \tan x,$$ $$1-\sin^2 x = \sin x$$ $$w^2+w-1=0, w =\sin x.$$ If $x_0$ is a root for it, then an answer is $cos(x_0)$.

Answer 2

$$x^2=\cos^2(\arctan(x))$$

Let $x=\tan(\theta)$

$$x^2+1=\sec^2(\theta)\implies \cos^2(\theta)=\frac{1}{1+x^2}$$

For the above, try to prove the solution must be positive.

$$x^2=\frac{1}{1+x^2}$$