Đáp án:
Giải thích các bước giải:
$\quad \tan^2x + \tan x.\tan3x = 2\qquad (*)$
$ĐKXĐ:\ \begin{cases}\cos x \ne 0\\\cos 3x \ne 0\end{cases}\Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{2} + n\pi\\x \ne \dfrac{\pi}{6} + \dfrac{n\pi}{3}\end{cases}$
$(*)\Leftrightarrow (1+ \tan^2x) + \dfrac{\sin x.\sin3x}{\cos x.\cos 3x}= 3$
$\Leftrightarrow \dfrac{1}{\cos^2x} + \dfrac{\cos2x - \cos4x}{\cos2x + \cos4x}= 3$
$\Leftrightarrow \dfrac{2}{1 + \cos2x} + \dfrac{\cos2x - 2\cos^22x + 1}{\cos2x + 2\cos^22x -1}= 3$
$\Leftrightarrow \dfrac{2}{1+\cos2x} - \dfrac{(2\cos2x +1)(\cos2x -1)}{(2\cos2x -1)(\cos2x -1)}= 3$
$\Leftrightarrow \dfrac{2(2\cos2x -1) - (2\cos2x +1)(\cos2x -1)}{(2\cos2x -1)(\cos2x +1)}= 3$
$\Leftrightarrow \dfrac{2\cos^22x - 5\cos2x +1}{(2\cos2x -1)(\cos2x +1)}= -3$
$\Leftrightarrow 2\cos^22x - 5\cos2x +1 = - 3(2\cos2x -1)(\cos2x +1)$
$\Leftrightarrow 4\cos^22x - \cos2x - 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos2x = \dfrac{1 +\sqrt{17}}{8}\\\cos2x= \dfrac{1 -\sqrt{17}}{8}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x = \arccos\left(\dfrac{1 +\sqrt{17}}{8}\right) + k2\pi\\2x = -\arccos\left(\dfrac{1 +\sqrt{17}}{8}\right) + k2\pi \\2x = \arccos\left(\dfrac{1 -\sqrt{17}}{8}\right) + k2\pi\\2x = -\arccos\left(\dfrac{1 -\sqrt{17}}{8}\right) + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac12\arccos\left(\dfrac{1 +\sqrt{17}}{8}\right) + k\pi\\x = -\dfrac12\arccos\left(\dfrac{1 +\sqrt{17}}{8}\right) + k\pi \\x = \dfrac12\arccos\left(\dfrac{1 -\sqrt{17}}{8}\right) + k\pi\\x = -\dfrac12\arccos\left(\dfrac{1 -\sqrt{17}}{8}\right) + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
