In post#11 Archie Meade reached the step

sinA + sinB = 2sin(A+B)/2*cos(A-B)/2

sinA - sinB = 2cos(A+B)/2*sin(A-B)/2.

From that step

(sinA + sinB)(sinA - sinB) = [2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]

sin^2A - sin^2B = sin(A+B)sin(A-B)

When you try to solve the equation, it is better to take A>B. So I sugested to write the equation as

sin^(2x) - sin^2(x-30) = 0