Question

Pls help i don’t know how to do this

Answer 1

Answer: $\displaystyle\\\frac{2\pi }{3},\ \frac{5\pi }{3} .$

Step-by-step explanation:

$\displaystyle\\tan\theta=-\sqrt{3}\ \ \ \ 0\leq \theta\leq 2\pi \\\theta =\frac{2\pi }{3}+\pi N\ \ \ (N=0,\ 1,\ 2,\ 3\ ...)\\ N=0\\\theta=\frac{2\pi }{3} +\pi *0\\\theta=\frac{2\pi }{3}\ \ \ \ ( 0\leq \theta\leq 2\pi)\\ N=1\\\theta=\frac{2\pi }{3}+\pi *1 \\\theta=\frac{2\pi }{3} +\pi \\\theta=\frac{2\pi +3\pi }{3} \\\theta=\frac{5\pi }{3} \ \ \ ( 0\leq \theta\leq 2\pi)$

$N=2\\\theta=\frac{2\pi }{3}+2\pi \\ \theta=\frac{2\pi +3*2\pi }{3}\\ \theta=\frac{2\pi +6\pi }{3} \\\theta=\frac{8\pi }{3} \\\theta=2\frac{2}{3} \pi \ \ \ \ (\notin0\leq \theta\leq 2\pi).$