Question

How do you solve the following equation `cos2x+cosx=0` in the interval [0, 2pi]?

Answer 1

I found:
`x=pi`
`x=pi/3 and (5pi)/3`

Explanation:

We can use trigonometric identities to change all into `cos` as:
`cos^2(x)-sin^2(x)+cos(x)=0`
and:
`cos^2(x)-1+cos^2(x)+cos(x)=0`
`2cos^2(x)+cos(x)-1=0`
We can solve this using the Quadratic Formula as in a second degree equation in `cos(x)`;
we can write `cos(x)=t`
and we get:
`2t^2+t-1=0`
`t_(1,2)=(-1+-sqrt(1+8))/4=(-1+-3)/4`
`t_1=-1`
`t_2=1/2`
but: `t=cos(x)` so we have:
`cos(x)=-1` when `x=pi`
`cos(x)=1/2` when `x=pi/3 and (5pi)/3`