Composition of $$\cos^{-1}x$$ with $$\cos$$

We know that

$$f(x)=\sin^{-1}(\sin x)$$

has range

$$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$

while

$$g(x)=\cos^{-1}(\cos x)$$

has range

$$[0,\pi].$$

Since

$$f(x)=g(x),$$

the common value must lie in

$$\left[0,\frac{\pi}{2}\right].$$

Let this common value be

$$t.$$

Then

$$\sin x=\sin t$$

and

$$\cos x=\cos t.$$

Since

$$t\in\left[0,\frac{\pi}{2}\right],$$

the simultaneous conditions imply

$$x=2n\pi+t,\qquad n\in\mathbb Z.$$

Now

$$f(2n\pi+t)=\sin^{-1}(\sin t)=t$$

and

$$g(2n\pi+t)=\cos^{-1}(\cos t)=t,$$

so every number of the form

$$x=2n\pi+t,\qquad 0\le t\le\frac{\pi}{2}$$

is a solution.

Restricting to

$$[-2\pi,2\pi],$$

we get:

For

$$n=-1,$$

$$x\in\left[-2\pi,-\frac{3\pi}{2}\right].$$

For

$$n=0,$$

$$x\in\left[0,\frac{\pi}{2}\right].$$

For

$$n=1,$$

$$x=2\pi$$

only.

Thus the solution set contains entire intervals, not just isolated points.

Therefore, there are infinitely many real solutions.

Hence, $$\text{Answer}=\text{Infinitely many}$$.