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

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You need to solve for `x in [0,2pi]` the equation `2cos x + 1 = 0` , such that:

`2cos x = -1 => cos x = -1/2 => {( cos x = cos(pi - pi/3)),( cos x = cos (pi + pi/3)):}`

`{(cos x = cos(2pi/3)),(cos x = cos (4pi/3)):} => {(x = (2pi)/3),(x = (4pi)/3):}`

**Hence, evaluating the solutions to the given equation, under the given conditions, yields `x = (2pi)/3` , `x = (4pi)/3.` **

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