Môn Toán Lớp 10 Tanx-cotx=3/2 ai giải pt này hộ em vs ạ
Question
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Trả lời ( )
$\tan x – \cot x = \dfrac{3}{2}$ $(*)$
$ĐK: \, \begin{cases}\sin x \ne 0\\\cos x \ne 0\end{cases} \Leftrightarrow x \ne k\dfrac{\pi}{2} \quad (k\in \Bbb Z)$
$(*) \Leftrightarrow -\dfrac{\cos^2x – \sin^2x}{\sin x\cos x} = \dfrac{3}{2}$
$\Leftrightarrow \dfrac{\cos2x}{\sin2x} = -\dfrac{3}{4}$
$\Leftrightarrow \tan2x = -\dfrac{4}{3}$
$\Leftrightarrow 2x = \arctan\left(-\dfrac{4}{3}\right) + k\pi$
$\Leftrightarrow x = \dfrac{1}{2}\arctan\left(-\dfrac{4}{3}\right) + k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
ĐK: $x\ne \dfrac{k\pi}{2}$
$\tan x-\cot x=\dfrac{3}{2}$
$\Leftrightarrow \dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}=\dfrac{3}{2}$
$\Leftrightarrow \dfrac{\sin^2x-\cos^2x}{\sin x\cos x}=\dfrac{3}{2}$
$\Leftrightarrow \dfrac{-\cos 2x}{\dfrac{1}{2}\sin2x}=\dfrac{3}{2}$
$\Leftrightarrow -2\cot2x=\dfrac{3}{2}$
$\Leftrightarrow \cot2x=\dfrac{-3}{4}$
$\Leftrightarrow x=\dfrac{1}{2}arccot\dfrac{-3}{4}+\dfrac{k\pi}{2}$ (TM)