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find all solutions:
3sin^2x+2sinx-1=0; in the interval [0,2pi)
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Introduce new variable u = sin(x).
Then your equation takes the form
3u^2 + 2u - 1 = 0.
Factorize
(3u-1)*(u+1) = 0.
Its roots are u = 1/3 and u= -1.
So, we consider now two cases
(a) sin(x) = 1/3, which gives two solutions x = and x = - .
(b) sin(x) = -1, which gives the solution x = = .
At this point, the problem is solved completely.
ANSWER. The solutions to the given equation in the given interval are , - and .
Solved.