Question

Solve for `x` in `sin2xsinx-cos2xcosx=-cosx`?

Answer 1

See below

Explanation:

Get everything in terms of cosine by applying the double angle formulas and Pythagorean identities:
`sin2xsinx-cos2xcosx=-cosx`
`(2sinxcosx)sinx-(2cos^2x-1)cosx=-cosx`
`2sin^2xcosx-(2cos^3x-cosx)=-cosx`
`2cosxsin^2x-(2cos^3x-cosx)=-cosx`
`2cosx(1-cos^2x)-(2cos^3x-cosx)=-cosx`
`2cosx-2cos^3x-2cos^3x+cosx=-cosx`
`-4cos^3x+4cosx=0`
`-4cos^3x+4cosx=0`
`-4cosx(cos^2x-1)=0`

Solutions:
`cosx=0`
`cosx=+-1`

In the interval `[0, 2pi)`
`x= pi/2, (3pi)/2, 0, pi`

Graph:
graph{sin(2x)sinx-cos(2x)cosx+cosx [-10, 10, -5, 5]}