ĐK: $\cos\left(x+\dfrac{\pi}{4}\right)\ne 0\to x\ne \dfrac{\pi}{4}+k\pi$
$1+\tan\left(x+\dfrac{\pi}{4}\right)=\cos^2x$
• Với $\cos x\ne 0$:
$\to 1+\dfrac{\tan x+1}{1-\tan x}=\cos^2x$
$\to \dfrac{1-\tan x+\tan x+1}{1-\tan x}=\cos^2x$
$\to \dfrac{2}{1-\tan x}=\cos^2x$
$\to 1-\tan x=2(1+\tan^2x)$
$\to 2\tan^2x+\tan x+1=0$
$\to $ VN, loại
• Với $\cos x=0\to x=\dfrac{\pi}{2}+k\pi$ (*)
$\tan\left(x+\dfrac{\pi}{4}\right)=-1$
$\to x=\dfrac{-\pi}{2}+k\pi$ (TM (*) )
Vậy $x=\dfrac{\pi}{2}+k\pi$
