My approach:

So,

Marla Payton
2021-12-30
Answered

Find the general solution of equation $\mathrm{cot}x+\mathrm{tan}x=2$

My approach:

$\mathrm{cot}x+\mathrm{tan}x=\frac{(1+{\mathrm{tan}}^{2}x)}{\mathrm{tan}x}=2$

${\mathrm{sec}}^{2}x+\mathrm{tan}x=2$

$\frac{1}{\left(\mathrm{sin}x\mathrm{cos}x\right)}=2$

${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=2\mathrm{sin}x\mathrm{cos}x$

${(\mathrm{sin}x-\mathrm{cos}x)}^{2}=0$

$\mathrm{sin}x=\mathrm{cos}x$

So,$x=n\pi +\left(\frac{\pi}{4}\right)$ . This was my answer. But the answer given is $x=2n\pi \pm \frac{\pi}{3}$

My approach:

So,

You can still ask an expert for help

Ben Owens

Answered 2021-12-31
Author has **27** answers

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

$\frac{1}{a}+a=2$

$1+{a}^{2}=2a$

${a}^{2}-2a+1=0$

${(a-1)}^{2}=0$

$a-1=0$

$a=1$

$\mathrm{tan}\left(x\right)=1$

Then it is

servidopolisxv

Answered 2022-01-01
Author has **27** answers

wrong reduction to algebraic form.

$\mathrm{tan}x\equiv u\Rightarrow u+\frac{1}{u}=2\Rightarrow u=1$

$u=1\Rightarrow x=n\pi +\frac{\pi}{4}$

Vasquez

Answered 2022-01-09
Author has **460** answers

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