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horrorfan [7]
1 year ago
9

Pls help i don’t know how to do this

Mathematics
1 answer:
algol [13]1 year ago
6 0

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).\\

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Subproblem 1:
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