Question

How do you find the general solutions for `cos 2x + cos x - 2 =0`?

Answer 1

`x = 2npi` for `AAn in ZZ`

Explanation:

`cos(2x) = 2cos^2(x)-1` `color(white)("XXXX")`(This is one version of the double angle formula for cos)

So
`color(white)("XXXX")` `cos(2x)+cos(x)-2 =0`
can be written as
`color(white)("XXXX")` `2cos^2(x)+cos(x)-3 = 0`

This can be factored as:
`color(white)("XXXX")` `(2cos(x)+3)(cos(x)-1) = 0`

So `cos(x) = -3/2` `color(white)("XXXX")`or`color(white)("XXXX")` `cos(x)=1`
But `-3/2` is not within the range of `cos(x)` for any value of `(x)`.
`color(white)("XXXX")`(`cos(x) in [-1,+1]`)

So `cos(x) = 1`
`color(white)("XXXX")`within the range `x in [0, 2pi)`
`x=0`
or, in general
`x=2npi` for `AA n in ZZ`