Đáp án:
$\begin{array}{l}
Dkxd:x \ne \dfrac{\pi }{2} + k\pi \\
1 + \tan x = 2\sqrt 2 \sin x\\
\Rightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = 2\sqrt 2 \sin x\\
\Rightarrow \cos x + \sin x = 2\sqrt 2 \sin x.\cos x\left( * \right)\\
\text{Đặt}:\cos x + \sin x = t\\
\Rightarrow {\left( {\cos x + \sin x} \right)^2} = {t^2}\\
\Rightarrow {\cos ^2}x + 2\cos x.\sin x + {\sin ^2}x = {t^2}\\
\Rightarrow 1 + 2.\cos x.sinx = {t^2}\\
\Rightarrow 2\cos x.\sin x = {t^2} - 1\\
\left( * \right) \Rightarrow t = \sqrt 2 .\left( {{t^2} - 1} \right)\\
\Rightarrow \sqrt 2 .{t^2} - t - \sqrt 2 = 0\\
\Rightarrow \left[ \begin{array}{l}
t = \sqrt 2 \\
t = - \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin x + \cos x = \sqrt 2 \\
\sin x + \cos x = - \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 \\
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{ - \sqrt 2 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
x + \dfrac{\pi }{4} = - \dfrac{\pi }{6} + k2\pi \\
x + \dfrac{\pi }{4} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{ - 7\pi }}{{12}} + 2k\pi \\
x = \dfrac{{11\pi }}{{12}} + k2\pi
\end{array} \right.\\
Do:x \in \left( {\pi ;2\pi } \right)\\
\Rightarrow x = \dfrac{{17\pi }}{{12}}\\
\Rightarrow B\\
15)\\
5\left( {\sin x + \cos x} \right) - 3.\sin x.\cos x - 3 = 0\\
\Rightarrow 5.t - 3.\dfrac{{{t^2} - 1}}{2} - 3 = 0\\
\Rightarrow 10t - 3{t^2} + 3 - 6 = 0\\
\Rightarrow 3{t^2} - 10t + 3 = 0\\
\Rightarrow \left( {3t - 1} \right)\left( {t - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = \dfrac{1}{3}\\
t = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{3}\\
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = 3\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{6}\\
\Rightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + \arcsin \left( {\dfrac{{\sqrt 2 }}{6}} \right) + k2\pi \\
x = - \dfrac{{3\pi }}{4} - \arcsin \left( {\dfrac{{\sqrt 2 }}{6}} \right) + k2\pi
\end{array} \right.
\end{array}$
