Решите уравнение:

1+cos (x/2) = 2sin ((x/4) - (3 п/2) ⋅2cos (2x/4) = - 2cos (x/4)

Cos^2x-cosx-2=0

обозн. cosx=t, |t|<=1

t2-t-2=0

d = (-1) ^2-4*1 * (-2) = 1+8=9

t1=1-3/2 t2=1+3/2

t1=-1 t2=2

t2>1

cosx=-1

x=pi+2pi*n

2.2cos^2x-sin4x=1

2 (1-sin^2x) - 2sin2xcos2x=1

2-2sin^2x-2 (2sinxcosx * (cos^2x-sin^2x) = 1

2-2sin^2x-4sinxcosx (cos^2x-sin^2x) - 1=0

(1-2sin^2x) - 4sinxcosx (1-sin^2x-sin^2x) = 0

(1-sin^2x) - 4sinxcosx (1-2sin^2x) = 0

(1-sin^2x) (1-4sinxcosx) = 0

1-sin^2x=0 или 1-4sinxcosx=0

sin^2x=1/2 1-2sin2x=0

x = (-1) ^n*arcsin (1/2) + pi*n sin2x=1/2

x = (-1) ^n*pi/6+pi*n 2x = (-1) ^n*arcsin (1/2) + pi*n

x = (-1) ^n*pi/6+pi*n x = (-1) ^n*pi/6+pi*n

x = (-1) ^n*pi/12+pi*n/2