Question

Find all solutions in the interval [0, 2π). sin2x + sin x = 0

Answer 1

\(\sin(2x)\) or \(\sin^2(x)\)?

Answer 2

sin2x=2sinxcosx
so
2sinxcosx+sinx=0
sinx(2cosx+1)=0
this will be 0 whne either sinx =0 or cosx=-1/2

Answer 3

sorry, its sin^2(x)

Answer 4

Find the roots of \(x^2+x=0\) and then solve for \(\sin(\theta)=x\)

Answer 5

ohh, then
make sinx=t
t^2+t=0
t(t+1)=0
so t=0, or t=-1
substituting back
sinx=0 or sinx=-1
means
x=0 or x=3Pi/2

Answer 6

@myko Is correct, but we mustn't forget that \(\sin(\pi)=0, \pi\in[0,2\pi)\)

Answer 7

so my answers will be 0, pi, and 3pi/2 then?

Answer 8

right. x=0 or Pi or x=3Pi/2

Answer 9

thanks very mcuh!!