Question #bf27d

First, use the identity:

`cos2x = cos^2x-sin^2x =color(red)(2cos^2x-1)`

Now we can substitute and solve this like a quadratic equation:

`cos2x + 3cosx = 1`
`color(red)(2cos^2x)+3cosx color(white)"-"color(red)(-1) = 1`
`2cos^2x + 3cosx-2=0`
`(2cosx-1)(cosx+2) = 0`

`cosx = 1/2` or `cosx=-2`

Since `cosx` cannot ever be `-2`, the only solutions are when `cosx = 1/2`.

Therefore, x could be `cos^-1(1/2) = pi/3`, or x could be `-pi/3` since cosine is an even function and therefore `cos(pi/3) = cos(-pi/3)`

Finally, `x` could also be `-pi/3` or `pi/3` plus any multiple of `2pi`, since adding `2pi` to an angle makes it go all the way around the circle and basically become the same angle again. So, our final answer is:

`{x | x=-pi/3+2kpi, x=pi/3+2kpi}` for any integer `k`.

`cos 2 x + 3 cos x =1`

`cos^2 x -sin^2 x + 3 cos x =1`, where `cos 2 x=cos^2 x -sin^2 x`

`cos^2 x -(1- cos^2 x )+ 3 cos x =1`, where `sin^2 x = 1 - cos^2 x`

`2 cos^2 x -1+ 3 cos x =1`

rearrange the equation,
`2 cos^2 x + 3 cos x - 2= 0`

`(2 cos x - 1)(cos x + 2) = 0`

`2 cos x -1 = 0, cos x = 1/2, x = cos^(-1)(1/2)`
`x = 60^@ , 300^@`

`cos x + 2 = 0, cos x = -2 ->` invalid