Q:

Trigonometric inequality with weird angle

I have this inequality $\log_{\tan x}\sqrt{\sin^2x-\frac{5}{12}}<1$ that I would like to solve.  
So $\tan x>0$ and $ \tan x\ne1$.  First I try when $\tan x>1$.
Then I have $\sqrt{\sin^2x-\frac{5}{12}}<\tan x$
I have to solve the system

$\sin^2x-\frac{5}{12}\ge0$ (I can't solve this one ($\frac{5}{12}$ troubles me I souldn't use a calculator or a table to solve this))
$\tan x>0$
$\sin^2x-\frac{5}{12}<\tan^2x$ and this one

A:

$1)$ If $\tan x>1\to \frac{\pi}{4}<x<\frac{\pi}{2}$ or $\frac{5\pi}{4}<x<\frac{3\pi}{2}$  $(1)$ then:
$$\sqrt{\sin^2x-\frac{5}{12}}<\tan x\to \sin^2x-\frac{5}{12}<\tan^2x\\
\sin^2x-\tan^2x<\frac{5}{12}\to -\frac{\sin^4x}{\cos^2 x}<\frac{5}{12}\\
-12\sin^4x<5(1-\sin^2x)\to12\sin^4x-5\sin^2x+5>0$$
which is true for any $x\in \Bbb R$ $(2)$. 
We also must have
$$\sin^2x\ge\frac{5}{12}\to\sin x\le-\frac{\sqrt{15}}{6} \text{ or } \sin x\ge\frac{\sqrt{15}}{6}\quad (3)$$
So, considering the intersection of $(1), (2)$ and $(3)$ we get the solution  
$$\frac{\pi}{4}<x<\frac{\pi}{2} \text{ or } \frac{5\pi}{4}<x<\frac{3\pi}{2}$$
Can you do the same idea for $0<\tan x <1$?

