# Logarithms with trigonometric inequality

My class is going to have an exam tomorrow, but we can't figure out how to solve such equations.

$$\log_{\ \large tg(x)} \sqrt{\sin(x)^2 - 5/12} < 1$$

We tried to transform $1$ to $\log_{\ \large tg(x)} tg(x)$ and solve it as

$$\sqrt{\sin(x)^2 - 5/12} < tg(x)$$

But we don't know how to continue. Any help will be greatly appreciated.

• What is $tg(x)$? Mar 16, 2014 at 21:41
• tangent [message too short] Mar 16, 2014 at 21:43
• What is $[a,b]$? An interval?
– MPW
Mar 16, 2014 at 21:52
• The step you have taken is only valid if $\log \tan x >0$. If it is negative, the inequality will reverse. And you must exclude the possibility that it is zero.
– MPW
Mar 16, 2014 at 21:55
• @MPW Someone just edited it, I wasn't sure how to express it. Mar 16, 2014 at 21:57

Square both sides, replace $\tan x$ with $\dfrac{\sin x}{\cos x}$ , and $\cos^2x$ with $1-\sin^2x$.