Well this is screaming out angle addition formulas. You see coscos - sinsin, what are the two arguments because you have the following identity:
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
--> in this case
cos(5x + 3x) = cos(5x)cos(3x) - sin(5x)sin(3x)
-->
cos(8x) = √3 / 2
--> solve cos(X) = √3/2
This is clearly a special right triangle, draw it out, find the angle that gives cosine equal to that, then go to your unit circle to make sure you get all solutions: http://sketchtoy.com/59497935
So we get: X = ±π/6...the -π/6 isn't going to be very helpful to us, so another solution set could be:
X = π/6, 11π/6 <-- 2π - π/6 = 12π/6 - π/6
Now we just find:
8x = X --> x = X/8 <-- need to be careful here. There is way more than just two solutions!
The general solution to the ORIGINAL problem is:
X = π/6 + 2πn or X = 11π/6 + 2πn
--> we need to keep finding values of n that gives X/8 within [0, 2π)
π/6 --> x = π/48
13π/6 --> x = 13π/48
25π/6 --> x = 25π/48
...you probably see the pattern, just add 12 each time...the same will happen for the other sequence
11π/6 --> x = 11π/48
23π/6 --> x = 23π/48
...
So this gives (I'll count two extra which will go over 2π...there should be 2*8 = 16 solutions):
x = π/48, 11π/48, 13π/48, 23π/48, 25π/48, 35π/48, 37π/48, 47π/48, 49π/48, 59π/48, 61π/48, 71π/48, 73π/48, 83π/48, 85π/48, 95π/48, 97π/48, 107π/48
Note that at 97π/48 it starts repeating: 97π/48 = 96π/48 + π/48 = 2π + π/48 <-- same angle as first solution. Likewise 107π/48 = 96π/48 + 11π/48 = 2π + 11π/48 <-- same as second solution.