# 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}

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.)