Number of solution of the equation $\cot^{-1}{\sqrt{4-x^2}+ \cos^{-1}{(x^2-5)}}=3π/2$

Question

Number of solution of the equation $ \cot^{-1}{\sqrt{4-x^2} + \cos^{-1}{(x^2-5)}}=3\pi/2$

$$ \cot^{-1}{\sqrt{4-x^2}+ \cos^{-1}{(x^2-5)}}={3π/2}$$
Taking sine both side and solving this is I get $$1 +\sqrt{5-x^2}x^2-4 \sqrt{5-x^2}+\sqrt{4-x^2}+\sqrt{x^6-15x^4+74x^2-120}=0$$ After this, I can't solve it and my approach is time taking also, so plz suggest me a simple approach and less time taking.

Answer 1

The expression requires $$ \begin{cases} x^2\le 4 \\[4px] |x^2-5|\le 1 \end{cases} $$ In particular, $x^2-5\ge-1$, that is, $x^2\ge 4$.

Answer 2

Using principal values of inverse trigonometric functions,

As $a=\sqrt{4-x^2}$ is defined and $\ge0$ for $4-x^2\ge0$

$\cot^{-1}a\le\dfrac\pi2$

and $\cos^{-1}(x^2-5)$ is defined and $\le\pi$ if $-\le x^2-5\le1$

So, the required equality will occur if $x^2-4=0$ and $x^2-5=-1$