Question 979386
I'm assuming you have {{{sin^2(x) + sin(x) = 0}}}


If {{{sin(x) = 0}}}, then it is true that {{{x = n*pi}}} where n is any integer. Ie, x is a multiple of pi.


If {{{sin(x) = -1}}}, then {{{x = (3pi)/2 + 2pi*n}}} where n is any integer. 


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Put the two together to get this solution set


{{{x = n*pi}}} or {{{x = (3pi)/2 + 2pi*n}}} where n is any integer.