Find general value of `theta` which satisfies both `sin theta = -1//2` and `tan theta=1//sqrt(3)` simultaneously.
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Here `sin theta lt 0` and `tan theta gt =0`, then `theta` lies in the third quadrant. Now `sin theta=-1/2` `rArr theta =pi+pi/6=(7pi)/6` Generalizing, we have `theta=2npi+(7pi)/6, n in Z`.
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