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

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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.

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**