Find the values of `theta` which satisfy `r sin theta=3` and `r=4 (1 + sin theta), 0 <= theta <= 2 pi`
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We have `r sin theta =3` and `r=4 (1+ sin theta)` Eliminating r, we get `4 sin^(2) theta +4 sin theta -3 =0` `:. (2 sin theta-1) (2 sin theta +3)=0` `rArr sin theta =1/2, -3/2` (not possible) `rArr theta=pi/6, (5 pi)/6` Thus, we have two solutions.
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