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)