Giải phương trình \(3\left( {\tan x + \cot x} \rig...

C

- Hướng dẫn giải

Lời giải chi tiết:

\(\begin{array}{l}3\left( {\tan x + \cot x} \right) = 2\left( {2 + \sin 2x} \right)\,\,\,\,\left( {x \ne \dfrac{{k\pi }}{2},\,\,k \in \mathbb{Z}} \right)\\ \Leftrightarrow 3\left( {\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}} \right) = 2\left( {2 + 2\sin x.\cos x} \right) \\ \Leftrightarrow 3.\dfrac{1}{{\sin x.\cos x}} = 4\left( {1 + \sin x.\cos x} \right)\end{array}\)

Đặt \(\sin x.cosx = t\,\,\,\,\left( {t \ne 0} \right)\), khi đó phương trình trở thành:

\(\begin{array}{l}3.\dfrac{1}{t} = 4\left( {1 + t} \right) \Leftrightarrow  - 4{t^2} - 4t + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{1}{2}\\t = \dfrac{{ - 3}}{2}\end{array} \right.\,\,\,\left( {tm} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\sin x.\cos x = \dfrac{1}{2}\\\sin x.\cos x = \dfrac{{ - 3}}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\dfrac{1}{2}\sin 2x = \dfrac{1}{2}\\\dfrac{1}{2}\sin 2x = \dfrac{{ - 3}}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = 1\\\sin 2x =  - 3\,\,\left( {loai} \right)\end{array} \right. \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\end{array}\)

Vậy \(S = \left\{ {\dfrac{\pi }{4} + k\pi \,,\,\,k \in \mathbb{Z}} \right\}\).