Solve the following equation. $${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0.$$

I can't figure out the way to solve this equation. This was my attempt \begin{array}{l} {\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0\\ \Leftrightarrow \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - 3(\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}}) = 0\\ \Leftrightarrow \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^2}x + {{\sin }^2}x}} - 3\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}} = 0\\ \Leftrightarrow \frac{{1 + 2{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\ \Leftrightarrow \frac{4}{{(1 - \cos 2x)(1 + \cos 2x)}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\ \Leftrightarrow \frac{4}{{1 - {{\cos }^2}2x}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0 \end{array} I don't know how to solve after that. Can anyone help me?


Let $\tan{x}-\cot{x}=t$.

Thus, $$t^2+2-3t=0.$$

For $t=1$ we obtain $$\tan^2x-\tan{x}-1=0,$$ which gives $x=\arctan\frac{1\pm\sqrt5}{2}+\pi k$, where $k\in\mathbb Z$.

For $t=2$ we obtain $$\tan^2x-2\tan{x}-1=0,$$ which gives $x=\arctan(1\pm\sqrt2)+\pi k$, where $k\in\mathbb Z$.

Actually, $\arctan(1+\sqrt2)=\frac{3\pi}{8}$ and $\arctan(1-\sqrt2)=-\frac{\pi}{8}$.

  • $\begingroup$ I got $$x=-\frac{\pi}{8}+k \frac{\pi}{2}$$ for $t=1$ and $$x = \frac{1}{2} \tan^{-1}\left( \frac{1}{2} \right)+\frac{\pi (2k-1)}{4}$$ for $t=2$. The solution period is $\frac{\pi}{2}$ and not $\pi$.I have confirmed that my solution indeed solves the original equation. $\endgroup$ – ja72 Sep 5 '17 at 17:00
  • $\begingroup$ For $t=2, $ $$1=\dfrac{2\tan x}{1-\tan^2x}=\tan2x$$ $\endgroup$ – lab bhattacharjee Sep 7 '17 at 15:16
  • $\begingroup$ @ja72 For $t=2$ they are the same. Draw the trigonometric circle. $\endgroup$ – Michael Rozenberg Sep 7 '17 at 15:21

$\tan^2 x + \cot^2x + 3(\tan x - \cot x) = 0 \implies (\tan x - \cot x)^2 + 2 + 3(\tan x - \cot x) = 0$. Now, let $y = \tan x - \cot x$, then $y^2-3y+2 = 0$, which gives $y=1 / y = 2$.

Hence, if we put $\tan x = z$, then $z - \frac1z = 1$ or $z - \frac 1z = 2$. The former simplifies to $z^2-z-1 = 0$ and the latter to $z^2-2z-1 =0$. Solve these quadratic equations, exclude the complex solutions, and the arctan of all remaining values of $z$ give you the possibilities for $x$.


As an alternative to Michael Rozenberg's answer, let $u=\tan x$. The equation becomes

$$u^2+{1\over u^2}-3\left(u-{1\over u}\right)=0$$

Multiplying through by $u^2$ to clear denominators, we wind up with


The quartic factors into two quadratics:


The rest follows as in Michael's answer.


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