Question

(cos3x+sin3x)^2=1+sin2x
Как это решать ?

Answer 1

(cos3x+sin3x)²=1+sin2x
cos²3x+2sin3xcos3x+sin²3x=1+sin2x
1+sin6x=1+sin2x
sin6x-sin2x=0
2sin(6x-2x)/2*cos(6x+2x)/2=0
2sin2x*cos4x=0
1)sin2x=0
2x=πk
x=πk/2
2)cos4x=0
4x=π/2+πk
x=π/8+πk/4;k€Z