How Do You Solve sinx(sin x + 1) = 0 for All Solutions?

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To solve the equation sinx(sin x + 1) = 0, identify the two cases: sin x = 0 and sin x = -1. For sin x = 0, the solutions are x = nπ, where n is any integer. For sin x = -1, the specific solution is x = -90 degrees or x = 270 degrees, which can also be expressed as x = 3π/2 + 2nπ for integer n. Remember that sine is a periodic function, so all solutions can be expressed in terms of these periodic properties. Understanding these periodic behaviors is crucial for finding all solutions.
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I'm having trouble starting off with these equations:

sinx(sin x + 1) = 0

I can solve sin x + 1 =
Sin x = -1
x = -90

but then I don't know how to find all the solutions. Can someone please explain it to me. Thanks.
 
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if h(x) = f(x)g(x), then h(x) = 0 if either f(x) = 0 or g(x) = 0, or f(x) = g(x) = 0, i.e. they have a common root.

Based on sinx (sin x + 1) = 0

one can have two possibilities: sin x = 0 and sin x = -1.

Consider if sin x = a, then x = sin-1 (a), and remember that sin is a periodic function, i.e. sin x = sin (pi - x) and sin x = sin (x + 2pi).
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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