Question

How do you solve `2 cos ^2 x - 5 cosx + 2 = 0`?

Answer 1

`x=+-pi/3 (+2npi, nin ZZ)`

Explanation:

Given
`color(white)("XXX")2cos^2x-5cos x + 2 =0`

Let `c=cos(x)`
then
`color(white)("XXX")2c^2-5c+2=0`

Factoring:
`color(white)("XXX")(c-2)(2c-1)=0`

Which implies
`color(white)("XXX")c=2`
or
`color(white)("XXX")c=1/2`

Since `c=cos(x)` and `cos(x)` has a Range of `[-1,+2]`
`color(white)("XXX")c=2` is an extraneous result

Therefore
`color(white)("XXX")c=cos(x)=1/2`
which is one of the standard angles with
`color(white)("XXX")x=+-pi/3` when `x in [0,2pi)`
or, in general
`color(white)("XXX")x=+-pi/3 +2npi, nin ZZ`