Assume the integers are degrees and not radians.
Use the identity cosx=sin(x+90)
Rewrite cos(Θ+10)=sin(2Θ-40) as
sin(Θ+100)=sin(2Θ-40)
The solution occurs at least where Θ+100=2Θ-40, or Θ=140.
sin140=sin(90+50)=sin(90-50)=sin40
In the first two quadrants 40, 140 work
cos50=sin40 and cos150=sin240
The remaining solutions are obtained by periodicity and staring at the graphs.
