Question

How do you find the general solutions for `cos(2x) + 5cos(x) + 3 = 0`?

Answer 1

For `x in [0,2pi]`
`color(white)("XXXX")` `x=pi`

Explanation:

`cos(2x) = 2cos^2(x)-1` `color(white)("XXXX")`(one of the double angle formulae)

`cos(2x)+5cos(x)+3=0`

`rArr` `color(white)("XXXX")` `2cos^2(x)+5cos(x)+3=0`

Factoring:
`rArr` `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`

Since `cos(x) in [-1,+1]` for all values of `x`
`color(white)("XXXX")` `cos(x)=-3/2` is an extraneous result.

`cos(x) = -1`
`rArr` `color(white)("XXXX")` `x= 3pi` within the range `[0,2pi]`

if the range is not restricted:
`color(white)("XXXX")` `x = pi+2npi, AAn in ZZ`

Note:
I modified a term in the question from (`cos2(x)`) to (`cos(2x)`);
the other possible interpretation might have been (`cos^2(x)`)
but that version has no solutions for `x`