solve for values of x for 0<= x < 2pi
sin^2 x + sin x = 0
we have : sin^2 x + sin x = 0
sin(x)(sin(x) + 1) = 0
so : sin(x) = 0 or sin(x) + 1 = 0
so : sin(x) = 0 => x= 0 + 2Kpi
and for : sin(x) + 1 = 0
so : sin(x) = -1 => sin(x) = sin(3pi/2) so : x = 3pi/2 + 2Kpi
and please if you have any other homework please invite me to i give you the best answer
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