Question

Find the solutions to the given equation in the interval [0,2π)?
2sin^3(x)−sin(x)cos(x)=2sin(x)

check my answers

π
4π)/3
2π/3

and would 0 be one of the answers? thanks

Answer 1

you are missing some solutions:
2sin^3(x)−sin(x)cos(x)=2sin(x)
2sin^3(x)−sin(x)cos(x) - 2sin(x) = 0
sinx(2sin^2 x - cosx - 2) = 0
sinx = 0 , x = 0 , 2π ---- in degrees : 0, 360°

or

2sin^2 x - cosx - 2 = 0
2(1 - cos^2 x) - cosx - 2 = 0
2 - 2cos^2 x - cosx - 2 = 0
2cos^2 x + cosx = 0
cosx(2cosx + 1) = 0

cosx = 0 , x = π/2, 3π/2 ---- in degrees : 90°, 270°
or
cosx = -1/2, x = 2π/3, 4π/3 ---- in degrees : 120°, 240°

x = 0, π/2, 2π/3, 4π/3, 3π/2 , (2π)

I am "old school" , not familiar with the interval notation of [0,2π), so
I think the 2π might be excluded, you decide.

to me something like 0 ≤ x < 2π was more obvious.

Answer 2

sinx = 0 , x = 0,π

Answer 3

Yup, been looking for that extra π ever since Thanksgiving.