Giải phương trình: tanx−3cotx=4(sinx+√3cosx)

Hướng dẫn giải:

Điều kiện:  \( \left\{ \begin{align} & \sin x\ne 0 \\  & \cos x\ne 0 \\ \end{align} \right.\Leftrightarrow \sin 2x\ne 0 \).

Lúc đó: (*) \( \Leftrightarrow \frac{\sin x}{\cos x}-3\frac{\cos x}{\sin x}=4\left( \sin x+\sqrt{3}\cos x \right) \)

 \( \Leftrightarrow {{\sin }^{2}}x-3{{\cos }^{2}}x=4\sin x\cos x\left( \sin x+\sqrt{3}\cos x \right) \)

 \( \Leftrightarrow \left( \sin x+\sqrt{3}\cos x \right)\left( \sin x-\sqrt{3}\cos x-2\sin 2x \right)=0 \)

 \( \Leftrightarrow \left[ \begin{align}  & \sin x=-\sqrt{3}\cos x \\  & \sin x-\sqrt{3}\cos x=2\sin 2x \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & \tan x=-\sqrt{3} \\  & \frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x=\sin 2x \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \tan x=\tan \left( -\frac{\pi }{3} \right) \\  & \sin \left( x-\frac{\pi }{3} \right)=\sin 2x \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{3}+k\pi  \\  & x-\frac{\pi }{3}=2x+k2\pi \vee x-\frac{\pi }{3}=\pi -2x+k2\pi  \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{3}+k\pi  \\  & x=-\frac{\pi }{3}-k2\pi \vee x=\frac{4\pi }{9}+\frac{k2\pi }{3} \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=-\frac{\pi }{3}+k\pi  \\  & x=\frac{4\pi }{9}+\frac{k2\pi }{3} \\ \end{align} \right.,\text{ }k\in \mathbb{Z} \) (nhận do  \( \sin 2x\ne 0 \)).