Help with one precalculus question

2cosx+1=0 for x

2cos x + 1 = 0

Subtract 1 from each side:

2cos x = -1

Divide each side by 2:

cos x = -1/2

Take the inverse cosine of both sides:

x = arccos(-1/2)

x = 120 degrees (2pi/3 radians)
also
x = 240 degrees (4pi/3 radians)

If domain of x is
zero to 2π

that is [0,2π)

then

2 cos x + 1 = 0

implies

2cos(x) = -1

so
.............-1
cos(x) =▬▬
..............2

and the solution values of x
are

2π/3 and 4π/3

If, however you need to solve for

all real solutions, then add 2π, a complete
revolution, times an integer to each of these solutions

x= 2π/3 +2kπ where k is any integer
and
x= 4π/3 +2kπ where k is any integer

It's not as hard as it looks babe :) .... follow these steps:
1)Subtract 1 from both sides you'll get: 2cosx= -1
2)Divide both sides by 2 you'll get: cosx= -1/2
3)Since you are solving for 'X' use the inverse of cos [arccos or cos^-1] you'll have cos^-1(-1/2) = 120 degrees or if you need it in radians than 2pi/3

That's all there is to it :) hope this helps

Easy as pi

2cosx+1=0
2cosx=-1
cosx=(-1/2)
x=arccos(-1/2)
x=120 degrees