Đáp án:
Bạn đang xem: 5tanx 2cotx 3=0
$\left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = \arctan\left(-\dfrac25\right) + k\pi\end{array}\right.\quad (\in\Bbb Z)$
Giải quí công việc giải:
$5\tan x -2\cot x - 3 = 0\qquad (*)$
$ĐK:\sin2x \ne 0\Leftrightarrow x \ne \dfrac{n\pi}{2}$
$(*)\Leftrightarrow 5\tan x -2\dfrac{1}{\tan x} - 3 = 0$
$\Leftrightarrow 5\tan^2x - 3\tan x - 2 = 0$
$\Leftrightarrow \left[\begin{array}{l}\tan x = 1\\\tan x =-\dfrac25\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = \arctan\left(-\dfrac25\right) + k\pi\end{array}\right.\quad (\in\Bbb Z)$
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