# 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.)

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Approved by eNotes Editorial Team

Posted on Approved by eNotes Editorial Team

Posted on Approved by eNotes Editorial Team

Posted on Approved by eNotes Editorial Team

Posted on