cos(2a) = cos^2(a) - sin^2(a)cos(2a) = 1 - 2sin^2(a)
Given that 4cos(a) = sin(2a), we can rewrite sin(2a) as 2sin(a)cos(a):
4cos(a) = 2sin(a)cos(a)2 = 2sin(a)sin(a) = 1
Therefore, the solution to the equation is a = π/2 + 2πn, where n is an integer.
cos(2a) = cos^2(a) - sin^2(a)
cos(2a) = 1 - 2sin^2(a)
Given that 4cos(a) = sin(2a), we can rewrite sin(2a) as 2sin(a)cos(a):
4cos(a) = 2sin(a)cos(a)
2 = 2sin(a)
sin(a) = 1
Therefore, the solution to the equation is a = π/2 + 2πn, where n is an integer.