Question

Solve:$(\cos x-\sin x)(2\tan x+\frac{1}{\cos x})+2=0$

Answer 1

$(\cos x-\sin x)(2\tan x+\frac{1}{\cos x})+2=0$

Let $t=\tan \frac{x}{2}$

$=\{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}-\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\}\{\frac{4\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}+\frac{1+{\tan }^2\frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\}+2=0$

$(\frac{1-t^2}{1+t^2}-\frac{2t}{1-t^2})(\frac{4t}{1-t^2}+\frac{1+t^2}{1-t^2})+2=0$

$\Rightarrow \frac{1-t^2-2t}{1+t^2}\times \frac{4t+1+t^2}{1-t^2}=0$

$\Rightarrow \frac{1-t^2-2t}{1+t^2}\times \frac{4t+1+t^2}{1-t^2}=0$

$\Rightarrow \frac{-(t^2+2t-1)(t^2+4t+1)}{1-t^4}=0$

$\Rightarrow -\frac{t^4+4t^3+t^2+2t^3+8t^2+2t-t^2-4t-1}{1-t^4}=0$

$\Rightarrow -\frac{t^4+6t^3+8t^2-2t-1}{1-t^4}=0$

$\Rightarrow \frac{t^4+6t^3+8t^2-2t-1}{1-t^4}=0$

$\Rightarrow \frac{t^4-1}{1-t^4}+\frac{2t(3t^2+4t-1)}{1-t^4}=0$

$\Rightarrow -1+\frac{2t(3t^2+4t-1)}{1-t^4}=0$

$\Rightarrow \frac{2t(3t^2+4t-1)}{1-t^4}=1$

$\Rightarrow 2t(3t^2+4t-1)=1-t^4$

$\Rightarrow 6t^3+8t^2-2t-1+t^4=0$

$\Rightarrow t^4+6t^3+8t^2-2t-1=0$

Let $t=\frac{\pm 1}{\sqrt{3}}\Rightarrow {\left(\frac{\pm 1}{\sqrt{3}}\right)}^4+6{\left(\frac{\pm 1}{\sqrt{3}}\right)}^3+8{\left(\frac{\pm 1}{\sqrt{3}}\right)}^2-2\left(\frac{\pm 1}{\sqrt{3}}\right)-1=0$

Hence its roots are $t_1=\frac{1}{\sqrt{3}}$ and $t_2=\frac{-1}{\sqrt{3}}$

$\Rightarrow \frac{x}{2}=n\pi \pm \frac{\pi }{6}$

$\Rightarrow x=2n\pi \pm \frac{\pi }{3}$ is required solution.