Question

How do you solve `cos^2 x + 2cosx=3` for x in the interval [0,2pi)?

Answer 1

`x=0`

Explanation:

`cos^2x+2cosx=3` can be written as `cos^2x+2cosx-3=0`

i.e. `cosx=(-2+-sqrt(2^2-4xx1xx(-3)))/2=(-2+-sqrt16)/2=(-2+-4)/2`

i.e.`cosx=1` or `cosx=-3`, but latter is out of range of `cosx`,

hence `cosx=1`

Hence `x=0`

as `2pi` is not included in `[0.2pi)`