Question

How do you find the general solutions for `4sin^3x=2sin^2x-2sinx-1=0`?

Answer 1

Solve: f(x) = 4sin^3x + 2sin ^2 x - 2sin x - 1 = 0

Explanation:

Call sin x = t -->
`4t^3 + 2t^2 - 2t - 1 = 0`
`2t^2 (2t + 1) - (2t + 1) = 0`
`(2t + 1) (2t^2 - 1) = (2t + 1)(sqrt2t - 1)(sqrt2t + 1) = 0`

a. `t = sin x = - 1/2` --> `x = -pi/6` and `x = -(5pi)/6`
b. `t = sin x = sqrt2/2` --> `x = pi/4` and `x = (3pi)/4`
c `t = sin x = -sqrt2/2` --> `x = -pi/4` and `x = - (3pi)/4`
Within the interval (0, 2pi), there are 6 answers:
`x = (7pi)/6 + 2kpi`
`x = (11pi)/6 + 2kpi`
`x = (pi)/4 + 2kpi`
`x = (3pi)/4 + 2kpi`
`x = (5pi)/4 + 2kpi`
`x = (7pi)/4 + 2kpi`