Find the general solution of equation \cot x+\tan x=2 My approach: \cot

Find the general solution of equation $\mathrm{cot}x+\mathrm{tan}x=2$
My approach:
$\mathrm{cot}x+\mathrm{tan}x=\frac{\left(1+{\mathrm{tan}}^{2}x\right)}{\mathrm{tan}x}=2$
${\mathrm{sec}}^{2}x+\mathrm{tan}x=2$
$\frac{1}{\left(\mathrm{sin}x\mathrm{cos}x\right)}=2$
${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=2\mathrm{sin}x\mathrm{cos}x$
${\left(\mathrm{sin}x-\mathrm{cos}x\right)}^{2}=0$
$\mathrm{sin}x=\mathrm{cos}x$
So, $x=n\pi +\left(\frac{\pi }{4}\right)$. This was my answer. But the answer given is $x=2n\pi ±\frac{\pi }{3}$
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Ben Owens
You know that $\mathrm{cot}\left(x\right)=\frac{1}{\mathrm{tan}\left(x\right)}$. Take random letter, let $a=\mathrm{tan}\left(x\right)$
Then it is
$\frac{1}{a}+a=2$
$1+{a}^{2}=2a$
${a}^{2}-2a+1=0$
${\left(a-1\right)}^{2}=0$
$a-1=0$
$a=1$
$\mathrm{tan}\left(x\right)=1$
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servidopolisxv
wrong reduction to algebraic form.
$\mathrm{tan}x\equiv u⇒u+\frac{1}{u}=2⇒u=1$
$u=1⇒x=n\pi +\frac{\pi }{4}$
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Vasquez

Note that it is enough to consider x>0. By AM-Gm inequality
$\frac{\mathrm{cot}x+\mathrm{tan}x}{2}\ge \sqrt{\mathrm{cot}x\mathrm{tan}x}=1$
for all $x\text{⧸}\in \left\{\frac{n\pi }{2}|n\in \mathbb{Z}\right\}Z$ and equality occur when
$\mathrm{cot}x=\mathrm{tan}x$
which implies $x\in \left\{n\pi +\frac{\pi }{4}|n\in \mathbb{Z}\right\}$

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