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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.