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

7. 2 sin^2x=sin x
Please answer asap

1 answer

2sin^2 x - sinx = 0
sinx(2sinx - 1) = 0
sinx = 0 or sinx = 1/2
x = 0,π, 2π
or
if sinx=1/2
x = π/6 or 5π/6
Similar Questions
    1. answers icon 0 answers
    1. answers icon 2 answers
  1. Find all solutions in the interval [0,2pi)4sin(x)cos(x)=1 2(2sinxcosx)=1 2sin2x=1 2x=1/2 x= pi/6, and 5pi/6 Then since its 2x i
    1. answers icon 1 answer
  2. cos^2(x) + sin(x) = 1- Find all solutions in the interval [0, 2pi) I got pi/2 but the answer says {0, pi/2, pi} Where does 0 and
    1. answers icon 2 answers
more similar questions