Question

Solve the equation over the interval [0,2pi)?

`2cos2x+2cosx=0`

Answer 1

The solutions are `x=pi/3,pi,(5pi)/3`

Explanation:

Use this identity:

`cos2x=2cos^2x-1`

Now, here's the actual problem:

`2cos2x+2cosx=0`

`2(color(red)(cos2x))+2cosx=0`

`2(color(red)(2cos^2x-1))+2cosx=0`

`color(red)(4cos^2x-2)+2cosx=0`

`4cos^2x+2cosx-2=0`

`4(cosx)^2+2cosx-2=0`

Substitute `u` for `cosx`:

`4u^2+2u-2=0`

`2u^2+u-1=0`

`(2u-1)(u+1)=0`

`u=1/2,-1`

Put `cosx` back in for `u`:

`cosx=1/2, qquadcosx=-1`

`x=pi/3,(5pi)/3, pi`

Hope this helped!

Answer 2

Use cos double angle identity and quadratic factoring
`x= pi/3, pi, (5pi)/3`

Explanation:

`2cos2x+2cosx=0`
`2(2cos^2x-1)+2cosx=0`
`4cos^2x-2+2cosx=0`
`4cos^2x+2cosx-2=0`
`2(2cos^2x+cosx-1)=0`
`2(2cos^2x+2cosx-cosx-1)=0`
`2(2cosx(cosx+1)-1(cosx+1))=0`
`2(2cosx-1)(cosx+1)=0`
`cosx =-1`
`cosx =1/2`
`x= pi/3, pi, (5pi)/3`