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

justaguide | Certified Educator

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)

=> x = 2*n*pi +- (2*pi)/3

The cosine function is symmetric about the y axis and we only require values of x in the set [0, 2*pi] .

So the values of x in the required set are (2*pi)/3 when n = 0; and (4*pi)/3 when n = 1.

The values of x in [0. 2*pi] for which 2cosx + 1 = 0 are (2*pi)/3 and (4*pi)/3 .

This can be verified from the graph of the cosine function.

ishpiro | Certified Educator

To solve the trigonometric equation 2cosx + 1 = 0 on the interval [0. 2pi],

first isolate the cosine, as follows:

2cosx = -1

cosx = -1/2

The general solution of the equation cosx = a, where -1 <a <1, is as follows:

x = +-arcos(a)+2pik, where k is an integer.

In our case, a = -1/2 and arcos(-1/2) = (2pi)/3 (Recall that the range of arccosine is [0, pi]). So the general solution of the given equation is as follows:

x = +-(2pi)/3 + 2pik, where k is an integer.

Two of the above values belong to the interval [0, 2pi]

when k = 0, x = 2pi/3  and

when k = 1, x = 2pi-(2pi)/3 = (4pi)/3.

So the values of on the interval [0, 2pi] satisfying the given equation are

(2pi)/3 and (4pi)/3 .

Neethu | Certified Educator

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}

hala718 | Certified Educator

2cosx + 1 = 0

First we will move 1 to the right side by subtracting 1 from both sides:

==> 2cosx + 1 - 1 = 0-1

==> 2cosx = -1

==> now we will divide by 2:

==> cosx = -1/2

Now we know that the angle whose cosine is 1/2 is ( pi/3)

But, since the sign is negative, then the angle is located in the second and third quadrants:

==> x1 = (pi - pi/3) = (2pi/3)

==> x2= (pi + pi/3) = 4pi/3

Then the solution is:

x= { 2pi/3 , 4pi/3 }

Here is a similar problem being solved in the interval 0 to 2pi.

giorgiana1976 | Student

We write 2cos x + 1 = 0 as:

cos x + cos x + 1 = 0

We'll substitute the values 1 = cos 0 and 0 = cos pi/2.

We'll subtract cos 0 both sides:

cos x + cos x = cos pi/2 - cos 2pi (1)

We'll transform the sum and the difference of cosines in products:

cos x + cos x = 2 cos [(x+x)/2]* cos [(x-x)/2]

cos x + cos x = 2cos x*cos 0 (2)

cos pi/2 - cos 2pi = 2 sin pi/4*sin (2pi- pi/2)/2

cos pi/2 - cos 2pi = 2sin pi/4*sin 3 pi/4 (3)

We'll substitute (2) and (3) in (1):

2cos x*cos 0 = 2sin pi/4*sin 3 pi/4

We'll divide by 2:

cos x*cos 0 = sin pi/4*sin 3 pi/4

We'll put cos 0 = 1

cos x*1= sin pi/4*sin 3 pi/4

sin pi/4 = cos (pi/2 - pi/4)

sin pi/4 = cos pi/4

cos x = cos pi/4

x = arccos (cos pi/4) + 2k*pi

x = pi/4

sin 3pi/4 = cos (pi/2 - 3pi/4)

x = cos (-pi/4)

Since the cosine function is even, we'll have:

cos (-pi/4) =cos (pi/4)

x = pi/4