Đáp án: $\left\{ \begin{array}{l}
x = \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}
\end{array} \right.$ $(k\in\mathbb Z)$
Giải thích các bước giải:
$\begin{array}{l}
1 + \tan x = 2\sqrt 2 \sin x\left( {\text{đkxđ: }\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi } \right)\\
\Rightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = 2\sqrt 2 \sin x\\
\Rightarrow \cos x + \sin x = 2\sqrt 2 \sin x.\cos x\\
\Rightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .2\sin x.\cos x\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 2x\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = 2x + k2\pi \\
x + \dfrac{\pi }{4} = \pi - 2x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {\text{tmđk}} \right)
\end{array}$
Vậy
$\left\{ \begin{array}{l}
x = \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}
\end{array} \right.$ $(k\in\mathbb Z)$
