Question

Need assistance with the question attached

Answer 1

Answer:

x = $\frac{4\pi }{9},\frac{10\pi }{9}$

Step-by-step explanation:

Given equation is,

$\sqrt{3}\text{tan}(\frac{2x}{3})+1=0$

By solving this equation,

$\text{tan}(\frac{2x}{3})=-\frac{1}{\sqrt{3}}$

$x=\frac{2}{3}\times \text{tan}^{-1}(-\frac{1}{\sqrt{3}})$ [tangent of any angle is negative in IInd and IVth quadrants]

x = $\frac{2}{3}\times (\frac{2\pi }{3})$, $\frac{2}{3}\times (\frac{5\pi}{3})$

x = $\frac{4\pi }{9},\frac{10\pi }{9}$

Therefore, value of x in the interval of [0, 2π] are $\frac{4\pi }{9},\frac{10\pi }{9}$.