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Mumz [18]
2 years ago
7

Need assistance with the question attached

Mathematics
1 answer:
natulia [17]2 years ago
6 0

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}.

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