Please help!!!!!!!!!!! Find all solutions in the interval [0,2π).
7. 2 sin^2x=sin x
Please answer asap
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
To find the solutions for the equation 2sin^2x = sinx in the interval [0,2π), follow these steps:
1. Write the equation in standard form: 2sin^2x - sinx = 0.
2. Factor out sinx: sinx(2sinx - 1) = 0.
3. Set each factor equal to zero and solve separately:
a) sinx = 0:
In the interval [0,2π), the solutions are x = 0 and x = π.
b) 2sinx - 1 = 0:
Solve for sinx: sinx = 1/2.
The unit circle shows that sinx = 1/2 at π/6 and 5π/6 in the interval [0,2π).
Therefore, the solutions for the equation 2sin^2x = sinx in the interval [0,2π) are: x = 0, x = π, x = π/6, and x = 5π/6.