Question

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

Answer 1

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