# What is root in equation 2(sin x)^3-cos2x-sinx=0? 0<= x<= 360 degrees

You need to write the equation in terms of `sin x` , hence, you need to write `cos 2x` in terms of sin x, using the following double angle identity, such that:

`cos 2x = 1 - 2sin^2 x`

Replacing `1 - 2sin^2 x` for `cos 2x` yields:

`2sin^3 x - (1 - 2sin^2 x) - sin x = 0`

`2sin^3 x - 1 + 2sin^2 x - sin x = 0`

You need to group the terms such that:

`(2sin^3 x + 2sin^2 x) - (1 + sin x) = 0`

Factoring out `2 sin^2 x` in the first group yields:

`2 sin^2 x(sin x + 1) - (sin x + 1) = 0`

Factoring out `sin x + 1 ` yields:

`(sin x + 1)(2 sin^2 x - 1) = 0`

Using the zero product rule, yields:

`sin x + 1 = 0 => sin x = -1 => x = (3pi)/2 `

`2 sin^2 x - 1 = 0 => sin^2 x = 1/2 => sin x = +-sqrt2/2`

`sin x = sqrt 2/2 => x = pi/4 or x = pi - pi/4 => x = (3pi)/4`

`sin x = -sqrt 2/2 => x = pi + pi/4 => x = (5pi)/4 or x = 2pi - pi/4 => x = (7pi)/4`

Hence, evaluating the solutions to the given equation, in `[0,2pi]` , yields `x = pi/4, x = (3pi)/4,x = (5pi)/4 ,x = (3pi)/2, x = (7pi)/4.`