Answer:
The question is given below as
[tex]\tan(3x)=-\sqrt{3}[/tex]Step 1:
Take the arctan of both sides
[tex]\begin{gathered} \tan(3x)=-\sqrt{3} \\ 3x=\tan^{-1}-\sqrt{3} \\ \tan^{-1}\sqrt{3}=60^0 \\ tan\text{ is \lparen-ve in the second and fourth quadrant\rparen} \\ hence \\ \theta=180-60^0=120^0 \\ \theta=360-60^0=300^0 \\ \frac{3x}{3}=\frac{120^0}{3},x=\frac{300}{3} \\ x=40^0,x=100 \\ x=\frac{2\pi}{9},x=\frac{5\pi}{9} \end{gathered}[/tex]Hence,
The values of x are given below as
[tex]x=\frac{2\pi}{9}+\frac{\pi n}{3}[/tex]where n could be, 1,2,3,4......