First, I will rewrite the question by replacing the x+2 portion with a different variable, alpha:
6cos alpha = 0
To solve this version, I will refer to the graph of cosine (or knowledge of how the function behaves for all alpha, to derive a general formula for the solution:
6 cos alpha = 0
cos alpha = 0
alpha = ..., -(5pi)/2, -(3pi)/2, -pi/2, pi/2, (3pi)/2, (5pi)/2, ...
:. alpha = pi/2 + npi, n in ZZ
However, keep in mind that we had used a variable substitution of alpha = x+2 to solve the original problem. Thus:
x + 2 = pi/2 + npi
x = pi/2 + npi - 2 = pi/2 - 2 + npi
For the original problem, we were asked to find all such solutions x for which 0 < x < 2pi. Therefore:
0 < x < 2pi
0 < pi/2 - 2 + npi < 2pi
2 - pi/2 < npi < 2pi - pi/2 + 2
2 - pi/2 < npi < (3pi)/2 + 2
0.429 < npi < 6.712color(white)("aaa")Approximately
From the final line, and knowing that pi ~~ 3.14159, it is evident that the only values for n which satisfy this relationship are n = 1 and n = 2. Thus:
x = (3pi)/2-2, (5pi)/2-2