99 баллов! Помогите решить! cos 4x+ 2cos^2 2x +4sin 2x=0

4cos^2 2x – 1 + 2cos^2 2x + 4sin2x = 0 4cos^2 2x + 4sin2x – 1 = 0 4 (1 - sin^2 2x) + 4sin2x – 1 = 0 4 – 4sin^2 2x + 4sin2x – 1 = 0 - 4sin^2 2x + 4sin2x + 3 = 0 4sin^2 2x -4sin2x – 3 = 0 sin2x = t 4t^2 -4t – 3 = 0 t_1,2 = (4±8)/8 t_1= 1,5 t_2 = - 1/2 sin2x = - ½ 2x = ( - 1) ^(n+1) arcsin (1/2) + pi n 2x= ( - 1) ^(n+1)* pi/6 + pi n x = ( - 1) ^(n+1)* pi/12 + (pi)/2 n