# 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

• You can remove the square root and write 1/2 before the log. – Jean-François Gagnon Mar 17 '17 at 19:37
• the square root is not a problem I get rid of the logarithm – yolo expectz Mar 17 '17 at 19:40

$1)$ 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 $0<\tan x <1$?

You are working too hard. Write: $$\log_{\tan x}\sqrt{\sin^2 x-\frac{5}{12}}=\frac{\ln\sqrt{\sin^2 x-\frac{5}{12}}}{\ln\tan x}$$ and analyse the functions.

• I don't really understand the natural $log$ – yolo expectz Mar 17 '17 at 20:24
• Don't you know how to express $\log_a(b)$ in terms of natural log? This is high-school stuff. – uniquesolution Mar 17 '17 at 23:15