Answer:
x = [tex]\frac{4\pi }{9},\frac{10\pi }{9}[/tex]
Step-by-step explanation:
Given equation is,
[tex]\sqrt{3}\text{tan}(\frac{2x}{3})+1=0[/tex]
By solving this equation,
[tex]\text{tan}(\frac{2x}{3})=-\frac{1}{\sqrt{3}}[/tex]
[tex]x=\frac{2}{3}\times \text{tan}^{-1}(-\frac{1}{\sqrt{3}})[/tex] [tangent of any angle is negative in IInd and IVth quadrants]
x = [tex]\frac{2}{3}\times (\frac{2\pi }{3})[/tex], [tex]\frac{2}{3}\times (\frac{5\pi}{3})[/tex]
x = [tex]\frac{4\pi }{9},\frac{10\pi }{9}[/tex]
Therefore, value of x in the interval of [0, 2π] are [tex]\frac{4\pi }{9},\frac{10\pi }{9}[/tex].
