$$\tan{x}+\cot{x}=8\cos{2x}$$ How to solve it with the simplest way? I managed to solve by changing both the $\tan{x}$ and $\cot{x}$ into $\cfrac{\sin{2x}}{1-\cos{2x}}$ and $\cfrac {1-\cos{2x}}{\sin{2x}}$. However, is there any easier way?

  • $\begingroup$ I take it that by "solve" you mean "Find $x$ such that ..."? The question would be a bit clearer if you said so. $\endgroup$ – David K Jul 24 '16 at 13:58
  • $\begingroup$ @Holmes, do you mean $\tan x+\cot x=8\cos2x$ or $\tan x+\cot x=\cos2x$? $\endgroup$ – ً ً Jul 24 '16 at 13:58
  • $\begingroup$ Have you tried using the double angle identities for cosine? $\endgroup$ – Brevan Ellefsen Jul 24 '16 at 14:19

$$\tan x + \cot x = \frac1{\sin x \cos x} = \frac1{\frac12 \sin(2x)}$$

So the equation is:

$$4\cos 2x \sin 2x = 1$$


$$\sin 4x = \frac12 = \sin (\pi/6)$$

Which has the solutions

  • $4x = \pi/6 \pmod {2\pi}$

  • $4x = \pi - \pi/6 \pmod {2\pi}$


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.