find all equations, sin(x)-2sin^2(x)=0?

Question

This should be easy, but I'm drawing a blank.

Answer 1

Factor out sin(x)
sin (x)[1-sin (x)]=0
Set each one equal to 0 and solve.
sin(x)=0, 1-sin (x)=0
x=0, x=pi/2

Answer 2

=sin x ( 1-2sin x) = 0
so sin x= 0 and x=0 and x=pi
1-2 sin x =0 sin x= 1/2 so x=pi6 and 5pi/6
taking only 0<=x<=2pi

Answer 3

Probably you mean find all solutions to
sin(x) - 2sin^2(x) = 0
sin(x) * (1 - 2sin(x)) = 0
So solutions are when sin(x) = 0
=> pi*k for any integer k

and sin(x) = 1/2
=> pi/2 +/- pi/3 + 2k*pi for any integer k

So total solutions are
{2kπ, 2kπ + π, 2kπ + π/6 , 2kπ + 5π/6} for all integers k

Answer 4

hm?
sin(x) ( 1 - 2 sin(x)) = 0
so either sin(x) = 0 which is 0, pi,
or 2 (sin(x)) = 1, which sin(x) = 1/2 pi/6