Question 716705
Since all multiples of {{{pi/6}}} are special angles and since the constant part of the argument, {{{5pi/6}}}, is already a multiple of {{{pi/6}}}, it will probably be easiest if we choose values for x that would make the other term in the argument, {{{(1/2)x}}}, also be a multiple of {{{pi/6}}}. By doing so we would then be able to add the fractions and end up with an argument that is also a multiple of {{{pi/6}}}.<br>
Perhaps you can already tell what values x should have that would turn {{{(1/2)x}}} into a multiple of {{{pi/6}}}. If not, then just pick a multiple of {{{pi/6}}}, set {{{(1/2)x}}} equal to it and solve for x. Let's be easy on ourselves and choose {{{pi/6}}} itself:
{{{(1/2)x = pi/6}}}
Multiplying both sides by 6 we get:
{{{3x = pi}}}
Dividing by 3 we get:
{{{x = pi/3}}}<br>
Perhaps you can now see that if x is any multiple of {{{pi/3}}}, then {{{(1/2)x}}} would end up being some multiple of {{{pi/6}}}. I'll get you started on a table of values:
<pre>
  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
</pre>I'll leave it up to you to find two more.<br>
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.