10cos(x)=-13+24sin(x), (rearranging)
100cos2(x)=(-13+24sin(x))2=169-624sin(x)+576sin2(x), (squaring both sides)
100-100sin2(x)=169-624sin(x)+576sin2(x),
676sin2(x)-624sin(x)+69=0,
sin(x)=(624±√(6242-4×676×69))/1352=(624±√202800)/1352,
sin(x)=0.124845 or 0.794625 approx.
x=7.38° or 52.62° approx. However, 7.38° doesn't satisfy the original equation. This result arose because of squaring earlier.
This angle can be expressed as 360n+52.62 degrees, where n is an integer.
However, there may be other solutions.
Consider rsin(A-B)=r(sinAcosB-cosAsinB)≡10cos(x)-24sin(x). B=x.
So rsinA=10, rcosA=24, from which r2=102+242=100+576=676, r=26.
tanA=10/24=5/12, A=22.62° approx.
Therefore, rsin(A-B)=26sin(22.62-x)=-13, sin(22.62-x)=-0.5.
22.62-x=-30° or -150°, making x=52.62° (as before) or 172.62°.
So the general answer is 360n+52.62 or 360n+172.62 degrees. Substituting these into the original equation confirms their validity.