Question

(sin3x+cos3x)^2=1+cos2x

Answer 1

1+SIN6X=1+COS2X

SIN6X=COS2X

COS2X-COS(п/2-6X)=0

-2SIN(п/4+4X)SIN(2X-п/4)=0

4X+п/4=пK

X=-п/16+пK/4

2X-п/4=пK

X=п/8+Пk/2