हल कीजिए -
`2 cos^(2) theta + 3 sin theta = 0`

Explanation

चूँकि `2 cos^(2) theta + 3 sin theta = 0`
`rArr 2(1-sin^(2) theta) +3 sin theta = 0`
`rArr -2 sin^(2) theta +3 sin theta + 2 = 0`
`rArr 2 sin ^(2) theta - 3 sin theta - 2 = 0`
`rArr 2 sin^(2) theta - 4 sin theta+ sin theta - 2 = 0`
`rArr (sin theta -2)(2 sin theta +1) = 0`
इसलिए , या तो `sin theta -2 = 0 " या " 2 sini theta + 1 = 0`
`rArr sin theta = 2 " या " sin theta = - 1/2`
`rArr sin theta = - 1/2" "(.:. sin theta = 2" सम्भव नहीं है ।")`
`rArr sin theta = - sin. pi/6 = sin (pi + pi/6)`
`rArr theta = (7pi)/6`
अतः व्यापक हल `theta = n pi + (-1)^(n) . (7pi)/6 , n in Z`