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

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justaguide eNotes educator| 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 eNotes educator| 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 eNotes educator| 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 eNotes educator| 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

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