2cosx + 1 = 0 find x values for the interval [0, 2pi]

We
have to find the value of x for `2cosx+1=0` in the interval `[0,2\pi]`

So,

`2cosx+1=0`

In other words,
`2cosx=-1`

`cosx=\frac{-1}{2}`

The solution will be of the
form:

`x=\pm cos^{-1}(\frac{-1}{2})+2\pi n` where n=0,1,2,. . .

For example, `x=\pm \frac{2\pi}{3} +2\pi n , \ n=0,1,2,...`

When
,

`n=0 : x=\pm \frac{2\pi}{3}`

`n=1:\ x=
\frac{2\pi}{3}+2\pi = \frac{8\pi}{3}`

or , 
`x=\frac{-2\pi}{3}+2\pi=\frac{4\pi}{3}`

However, we have to find the angle in
the interval `[0,2\pi]`

As a result, the answer is:

`x=\frac{2\pi}{3} \ and \ x=\frac{4\pi}{3}`