Sin^2x+sinx=0

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Let's solve the trigonometric equation ( \sin^2(x) + \sin(x) = 0 ):

We notice that both terms have a common factor of sin(x), so we can factor out sin(x) from the equation:

( \sin(x) (\sin(x) + 1) = 0 )

Now, we have two possibilities:

1) ( \sin(x) = 0 )

This occurs when x is a multiple of π:

( x = n\pi ) where n is an integer.

2) ( \sin(x) + 1 = 0 )

This implies that ( \sin(x) = -1 ), which occurs when x is in the form:

( x = (2n+1) \frac{\pi}{2} ) where n is an integer.

Therefore, the solutions to the equation ( \sin^2(x) + \sin(x) = 0 ) are:

( x = n\pi ) and ( x = (2n+1) \frac{\pi}{2} ) where n is an integer.