Question

How do you solve `2cos2x + 2cosx-1=0`?

Answer 1

`2cos2x+2cosx−1=0` implies `x = arccos( (sqrt(13) - 1)/4)`
(Yes; it is ugly... but the point is how to get there...)

Remember the double angle formula
`cos(2x) = 2 cos^2(x) - 1`

So
`2cos^2x+2cosx−1=0`
becomes
`4 cos^2 -2 +2 cos x - 1 = 0`
`4 cos^2 x + 2 cos x - 3 = 0`

Using the standard formula for roots of a quadratic:
`(-b +- sqrt( b^2 - 4ac))/2a`

We get (with minor handling)
`cos(x) = ( (-1) +- sqrt(13))/4`

`x = arccos ( cos(x) )` by definition
noting, however, that cos(x) must fall in the range [-1, +1]
so the root `( (-1) - sqrt(13))/4` can be eliminated

Therefore
`x = arccos(( (-1) + sqrt(13))/4)`