Find all values of x in the interval [0, 2pi] that satisfy the equation/inequality.

Find all values of x in the interval [0, 2pi] that satisfy 2 cos x + 1 > 0.

With no parentheses.

I got:
0 \< x < (2pi / 3), (4pi / 3) < x < 2pi.
Is this correct?

Also:
Find all values of x in the interval [0, 2pi] that satisfy 2 cos x - 1 = 0.

I got (pi / 3), (5pi / 3). The answers listed in the book are the same as mine.

P.S. Is there any reason why the 1 would be added to both sides of the equation first, but the 1 not subtracted from both sides of the inequality first?

1.

2cos(x) + 1 > 0

2cos(x) > -1

cos(x) > -1/2

x > 2π/3

And

x > 4π/3

2.

2cos(x) - 1 = 0

2cos(x) = 1

cos(x) = 1/2

x = π/3

And

x = 5π/3