Решите уравнение (подробно) 10 класс

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

2cos^2x-1-sin4x=0

cos2x-2sin2xcos2x=0

cos2x=0

x=П/4+Пn/2

1-2sin2x=0

sin2x=1/2

x=П/12+Пk

x=5 П/12+Пk