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Given:
[tex]sin^2x+sinx=0[/tex]To solve the function, factor the function using sin(x):
[tex]sin(x)*[sin(x)+1\rbrack=0[/tex]That means, in order to the produtct be zero, or sin(x) or [sin(x)+ 1] = 0
So, let's solve each situation.
[tex]\begin{gathered} sin(x)=0 \\ By\text{ taking the inverse:} \\ x=sin^{\left\{-1\right\}}\left(0\right) \\ x=\pi n,\text{ n is a integer} \end{gathered}[/tex]Or:
[tex]\begin{gathered} sin(x)+1=0 \\ \text{ Subtracting 1 from both sides:} \\ sin(x)+1-1=0-1 \\ sin(x)=-1 \\ \text{ Taking the inverse:} \\ x=sin^{-1}(-1) \\ x=\frac{3}{2}\pi+2\pi n \end{gathered}[/tex]Answer:
x = πn, 3/2 π + 2πn, for n integer n.