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

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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}`

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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)`

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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