if $a$ is are real number that $a \neq 0$, and $\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$, $x$ is on which trigonometric quadrants? 
if $a$ is are real number that $a \neq  0$, and $\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$, $x$ is on which trigonometric quadrants?

Things I have done so far: this problem is mostly different from that I previously solved.My Idea was to powering up both sides,take all to one side and then solve it like equation.which I was not successful. any starting hint would be appreciated as I don't looking for full solution.
UPDATE
thanks to ganeshie8 hints,i reached this right now$$\cos^2a=1+\sin a\times \cos a$$ 
And I stuck here.
 A: The equation has the first quadrant solution $x=\frac{\pi}{2}$. Apart from that,  the first quadrant is not possible. Note that if $x\ne \frac{\pi}{2}$, and $x$ is in the first quadrant, and $\cot x$ is defined, then $\cot x\gt 0$.
Thus if $\cot x\gt a^2$, then $0\lt \cot x-a^2\lt \cot x$, and therefore  $\frac{\cot x}{\cot x-a^2}\gt 1$. So our square root is greater than $1$, which is impossible for a cosine. 
And if $a^2\gt \cot x$, we are taking the square root of a negative number.  
A: We are given: 
$$\cos x = \sqrt{\frac{\cot x}{\cot x - a^2}}$$
For this to be true, 
$$\sqrt{\frac{\cot x}{\cot x - a^2}} \in [-1,1]$$
Now, for any$\sqrt{\frac{a}{b}} \in \mathbb R, \quad a>0 ,\quad b>0$
For the first condition ($a>0$),
$$ \implies \cot x > 0
\implies \frac{\cos x}{\sin x} >0\\
\implies (2n-1)\frac{\pi}{2}<x<(2n+1)\frac{\pi}{2} \text{ and } x \ne 0 \quad \forall \space n \in \mathbb Z$$
This means that $x$ is in  Quadrants I or IV.
For the second condition ($b>0$),
$$\implies \cot x - a^2 > 0 $$
We can now clearly see that 
$$\cot x > \cot x - a^2 > 0\\
\implies \text{Numerator} > \text{Denominator}\\
\implies \frac{\cot x}{\cot x - a^2} > 1$$
But this is against our initial assumption.
The only solution is $\frac{\pi}{2} \space\implies x \in $ Ist Quadrant 
