Giải phương trình $\tan \left(\frac{4\pi }{9}+x\right)+2\cot \left(\frac{\pi }{18}-x\right)=\sqrt{3}$ .

Điều kiện

$\left\{\begin{array}{l}\cos \left(\frac{4\pi }{9}+x\right)\ne 0\\ \sin \left(\frac{\pi }{18}-x\right)\ne 0\end{array}\right.\iff \left\{\begin{array}{l}\frac{4\pi }{9}+x\ne \frac{\pi }{2}+k\pi \\ \frac{\pi }{18}-x\ne k\pi \end{array}\right.\iff \left\{\begin{array}{l}x\ne \frac{\pi }{18}+k\pi \\ x\ne \frac{\pi }{18}-k\pi \end{array}\right.\iff x\ne \frac{\pi }{18}+k\pi$  , $\left(k;m\in \mathbb{Z}\right)$ .

Ta có $\left(\frac{4\pi }{9}+x\right)+\left(\frac{\pi }{18}-x\right)=\frac{\pi }{2}\Rightarrow \tan \left(\frac{4\pi }{9}+x\right)=\cot \left(\frac{\pi }{18}-x\right)$ .

$\left(2\right)\iff \cot \left(\frac{\pi }{18}-x\right)+2\cot \left(\frac{\pi }{18}-x\right)=\sqrt{3}\iff 3\cot \left(\frac{\pi }{18}-x\right)=\sqrt{3}$

 

$\iff \cot \left(\frac{\pi }{18}-x\right)=\frac{\sqrt{3}}{3}\iff \frac{\pi }{18}-x=\frac{\pi }{3}+k\pi \iff x=-\frac{5\pi }{18}-k\pi$

Vậy phương trình đã cho có nghiệm là $x=-\frac{5\pi }{18}+k\pi$ , $\left(k\in \mathbb{Z}\right)$.