Question

Find all solutions in the interval [0, 2π).

sin^2 x + sin x = 0

Answer 1

\(\color{#000000 }{ \displaystyle \sin^2x+\sin x = 0}\)

\(\color{#000000 }{ \displaystyle \sin x(\sin x +1)= 0}\)

will divide both sides by \(\color{#000000 }{ \displaystyle \sin x }\), and we get.

\(\color{#000000 }{ \displaystyle \sin x+1 = 0}\)

and note that when we divide by \(\color{#000000 }{ \displaystyle \sin x }\),
then \(\color{#000000 }{ \displaystyle \sin x = 0}\) is also a solution.

Answer 2

So you are going to have to equations to solve,
\(\color{#000000 }{ \displaystyle \sin x = 0}\) and \(\color{#000000 }{ \displaystyle \sin x +1= 0}\)

Answer 3

Okay that makes sense. How would I go about solving these equations?

Answer 4

Well, you can refer to the unit circle, and tell me for what values of x, will sin(x) be equal to zero?

Answer 5

pi?

Answer 6

Then for the next one it would be sin x = -1 so the answer would be 3pi/2?

Answer 7

@SolomonZelman

Answer 8

Yes, for sin(x)=0,
x=π, and another solution is x=0.
(0 is in the interval)

And for sin(x)=-1
you also got the correct solution,
x=3π/2

Answer 9

Okay, thank you very much!