Write the left side in terms of the sum and difference of 4x and x.
sin(4x +x) - sin(4x -x) = sin(x)
(sin(4x)cos(x) +cos(4x)sin(x)) - (sin(4x)cos(x) -cos(4x)sin(x)) = sin(x)
2cos(4x)sin(x) = sin(x)
sin(x)·(2cos(4x) -1) = 0 . . . . . subtract sin(x) and factor
Now, the zero product rule comes into play. Solutions are found where the factors are zero.
... sin(x) = 0 at x ∈ {0, 180°}
... 2cos(4x) -1 = 0 at x = arccos(1/2)/4 = (±60° +k·360°)/4 . . . . k an integer
... ... x ∈ {15°, 75°, 105°, 165°}