Sin2x+cos2x+tanx=2 HLep me

Điều kiện: $\cos x\ne 0\Rightarrow x\ne \dfrac{\pi} 2+k\pi$

$\begin{array}{l}
\tan x = t\\
 \Rightarrow \sin 2x = \dfrac{{2t}}{{1 + {t^2}}},\cos 2x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}\\
\sin 2x + \cos 2x + \tan x = 2\\
 \Leftrightarrow \dfrac{{2t}}{{1 + {t^2}}} + \dfrac{{1 - {t^2}}}{{1 + {t^2}}} + t = 2\\
 \Leftrightarrow 2t + 1 - {t^2} + t\left( {{t^2} + 1} \right) = 2\left( {{t^2} + 1} \right)\\
 \Leftrightarrow 2t + 1 - {t^2} + {t^3} + t = 2{t^2} + 2\\
 \Leftrightarrow {t^3} - 3{t^2} + 3t - 1 = 0\\
 \Leftrightarrow {\left( {t - 1} \right)^3} = 0\\
 \Leftrightarrow t = 1 \Leftrightarrow \tan x = 1\\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}$