Ok... I guess I'm being really stupid or something.
Everything below is in degrees.
\(\sin(x) = \cos(x -90)\\ \sin(x-30) = \cos(x-120)\\ \text{we also have }\cos(x) = \cos(-x) \\ \text{and }\cos(2x + 360k)=\cos(2x)\)
as we move forward from x=0 the first place the equation is satisfied is at
\(\cos(2x) = \cos(120-x)\\ 3x=120\\ x=40\)
the second place the equation is satisfied is at
\(\cos(2x-360)=\cos(120-x)\\ 2x-360 = 120-x\\ 3x = 480 \\ x=160\)
the third place the equation is satisfied is at
\(\cos(2x-360) = \cos(x-120)\\ 2x-360=x-120\\ x=240\)
and the final place less than 2pi is at
\(\cos(720-2x) = \cos(x-120)\\ 720-2x=x-120\\ 840=3x\\ x = 280\)
So the solutions, as Melody showed graphically, are
x = 40, 160, 240, 280 degrees.
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