Sina =cos2a, i.e.
sina= 1-2(sina)^2,
Let t=sina, then t∈[- 1, 1]
T =1-2t 2, i.e.
2t^2+t- 1=0
(2t- 1)(t+ 1)=0
The solution is t= 1/2 or t=- 1, which satisfies t∈[- 1, 1].
To sum up, sina= 1/2 or sina=- 1, that is
A=π/6 or a=-π/2 (other solutions with multiple cycles in one cycle)