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#
