Question

How do you solve `sin^2xcosx=cosx`?

Answer 1

`pi/2,, (3pi)/2`

Explanation:

Bring the equation to standard form:
`sin^2 x.cos x - cos x = 0`
`cos x(sin^2 x - 1) = 0`

a. cos x = 0 --> Unit circle -->
`x = pi/2`, and `x = (3pi)/2`
b. `sin^2 x - 1 = 0` --> `sin^2 x = 1`
`sin x = +- 1`
1. sin x = - 1 --> `x = (3pi)/2`
2. sin x = 1 --> `x = pi/2`

Answers for `(0, 2pi)`:
`pi/2, (3pi)/2,`