Find all solutions in the interval [0, 2π).
sin2 x + sin x = 0

Question

Find all solutions in the interval [0, 2π). 
sin2 x + sin x = 0

kirbykirby · Answer

Is that $\sin^2x+\sin x=0$? If so, a great big hint is to try substituting $y=\sin x$, which will give you a "normal" looking quadratic equation. Solve that, then re-substitute back the $y$'s with $\sin x$.

anonymous · Answer

yes it is that! @kirbykirby

anonymous · Answer

x = 0, π, 3pi/2
	x = 0, π, pi/3, 5pi/3
	x = 0, π,  pi/3,2pi/3
	x = 0, π,  4pi/3, 5pi/3

anonymous · Answer

those were pi signsand when i did the problem i got C but im not sure if im correct @kirbykirby

kirbykirby · Answer

What did you get as solutions to the quadratic equation if you substitute $\sin x$ with another variable?

anonymous · Answer

to be honest i made a guess but Im not sure how I came up with this conclusion because Ive done the problem once before but I never knew if i got it right lol @kirbykirby

kirbykirby · Answer

Hm ok well we'll try and go through this together. If you try this substitution:  $y=\sin x$, you end up with 
$y^2+y=0$
Factoring gives:
$y(y+1)=0$

So What are the solutions for $y$?

anonymous · Answer

y=0,-1?

anonymous · Answer

thats what I got when i factored

kirbykirby · Answer

Yep that's great , now if we substitute back $y = \sin x $, you get
$\sin x = 0$ and 
$\sin x =-1$

kirbykirby · Answer

Now you need to find the x-values that satisfy sin x = 0, and sin x = -1

kirbykirby · Answer

You can use a unit circle for that

anonymous · Answer

:/ not understanding how to get the solutions @kirbykirby

kirbykirby · Answer

Well, in the unit circle, you have the coordinates expressed as (cos x, sin x)

And the main points of interest on the circle are those that gives values of 0 and -1 for sine:

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