Since all multiples of are special angles and since the constant part of the argument, , is already a multiple of , it will probably be easiest if we choose values for x that would make the other term in the argument, , also be a multiple of . By doing so we would then be able to add the fractions and end up with an argument that is also a multiple of .
Perhaps you can already tell what values x should have that would turn into a multiple of . If not, then just pick a multiple of , set equal to it and solve for x. Let's be easy on ourselves and choose itself:
Multiplying both sides by 6 we get:
Dividing by 3 we get:
Perhaps you can now see that if x is any multiple of , then would end up being some multiple of . I'll get you started on a table of values:
x (1/2)x (1/2)x+5pi/6 sin((1/2)x+5pi/6) 2sin((1/2)x+5pi/6) f(x)
pi/3 pi/6 6pi/6 = pi 0 0 0
-pi/3 -pi/6 4pi/6 = 2pi/3 sqrt(3)/2 sqrt(3) sqrt(3)
0 0 5pi/6 1/2 1 1
I'll leave it up to you to find two more.
P.S. We did not have to make x be a value that made the argument of sin be a multiple of pi/6. We just did that because it made things easier. Any x that would make (1/2)x+5pi/6 turn out to be any of the special angles would work.