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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.

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  • $\begingroup$ What is $tg(x)$? $\endgroup$ – Priyatham Mar 16 '14 at 21:41
  • $\begingroup$ tangent [message too short] $\endgroup$ – Deepsy Mar 16 '14 at 21:43
  • $\begingroup$ What is $[a,b]$? An interval? $\endgroup$ – MPW Mar 16 '14 at 21:52
  • $\begingroup$ 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. $\endgroup$ – MPW Mar 16 '14 at 21:55
  • $\begingroup$ @MPW Someone just edited it, I wasn't sure how to express it. $\endgroup$ – Deepsy Mar 16 '14 at 21:57
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Square both sides, replace $\tan x$ with $\dfrac{\sin x}{\cos x}$ , and $\cos^2x$ with $1-\sin^2x$.

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