Question

Pls help i don’t know how to do this

Answer 1

Answer:
`\displaystyle\\(2\pi )/(3),\ (5\pi )/(3) .`

Explanation:

`\displaystyle\\tan\theta=-√(3)\ \ \ \ 0\leq \theta\leq 2\pi \\\theta =(2\pi )/(3)+\pi N\ \ \ (N=0,\ 1,\ 2,\ 3\ ...)\\ N=0\\\theta=(2\pi )/(3) +\pi *0\\\theta=(2\pi )/(3)\ \ \ \ ( 0\leq \theta\leq 2\pi)\\ N=1\\\theta=(2\pi )/(3)+\pi *1 \\\theta=(2\pi )/(3) +\pi \\\theta=(2\pi +3\pi )/(3) \\\theta=(5\pi )/(3) \ \ \ ( 0\leq \theta\leq 2\pi)\\`

`N=2\\\theta=(2\pi )/(3)+2\pi \\ \theta=(2\pi +3*2\pi )/(3)\\ \theta=(2\pi +6\pi )/(3) \\\theta=(8\pi )/(3) \\\theta=2(2)/(3) \pi \ \ \ \ (\\otin0\leq \theta\leq 2\pi).\\`