Giải phương trình: \(\tan x - 3\cot x = 4\left( {\...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\tan x - 3\cot x = 4\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow \frac{{\sin x}}{{\cos x}} - 3\frac{{\cos x}}{{\sin x}} = 4\left( {\sin x + \sqrt 3 \cos x} \right)\,\,\,\left( 1 \right)\end{array}\)

ĐK:  \(\left\{ \begin{array}{l}\cos x \ne 0\\\sin x \ne 0\end{array} \right. \Leftrightarrow x \ne \frac{{k\pi }}{2}\,\,\left( * \right)\).

\(\begin{array}{l}\left( 1 \right) \Leftrightarrow \frac{{{{\sin }^2}x - 3.{{\cos }^2}x}}{{\sin x.cosx}} = 4.\left( {sinx + \sqrt 3 .cosx} \right)\\{\rm{    }} \Leftrightarrow \left( {sinx - \sqrt 3 .cosx} \right)\left( {sinx + \sqrt 3 .cosx} \right) = 4.\sin x.cosx.\left( {sinx + \sqrt 3 .cosx} \right)\\{\rm{    }} \Leftrightarrow \left( {sinx + \sqrt 3 .cosx} \right).\left[ {sinx - \sqrt 3 .cosx - 2\sin 2x} \right] = 0\\{\rm{    }} \Leftrightarrow \left[ \begin{array}{l}\sin x + \sqrt 3 .\cos x = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\\sinx - \sqrt 3 .cosx - 2\sin 2x = 0\,\,\left( 3 \right)\end{array} \right.\end{array}\)

\(\begin{array}{l}\left( 2 \right) \Leftrightarrow \frac{1}{2}.\sin x + \frac{{\sqrt 3 }}{2}.\cos x = 0 \Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = 0\\{\rm{    }} \Leftrightarrow x + \frac{\pi }{3} = k\pi  \Leftrightarrow x =  - \frac{\pi }{3} + k\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}\left( 3 \right) \Leftrightarrow sinx - \sqrt 3 .\cos x = 2.\sin 2x \Leftrightarrow \frac{1}{2}.\sin x - \frac{{\sqrt 3 }}{2}.\cos x = \sin 2x\\{\rm{    }} \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 2x \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{3} = 2x + k2\pi \\x - \frac{\pi }{3} = \pi  - 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{3} + k2\pi \\x = \frac{{4\pi }}{9} + \frac{{k2\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

KL: \(x \in \left\{ { - \frac{\pi }{3} + k2\pi ;\frac{{4\pi }}{9} + \frac{{k2\pi }}{3};k \in \mathbb{Z}} \right\}\).

Chọn A.