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}`

The equation 2cosx + 1 = 0 has to be solved for values of x in the set `[0,2pi]`

`2*cos x + 1 = 0`

=> `2*cos x = -1`

=> `cos x = -1/2`

=> `x = 2*n*pi +-cos^-1(-1/2)`

(The entire section contains 4 answers and 390 words.)

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