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$$\tan{x}+\cot{x}=8\cos{2x}$$ 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?

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

So the equation is:

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

or

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

Which has the solutions

  • $4x = \pi/6 \pmod {2\pi}$

  • $4x = \pi - \pi/6 \pmod {2\pi}$

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