Question

$1+sin2x=(cos3x+sin3x)^{2}$

Answer 1

1 + Sin2x = (Cos3x + Sin3x)²

1 + Sin2x = Cos²3x + 2Sin3xCos3x + Sin²3x

1 + Sin2x = 1 + Sin6x

Sin6x - Sin2x = 0

$2Sin\frac{6x-2x}{2} Cos\frac{6x+2x}{2}=0\\\\Sin2xCos4x=0\\\\1)Sin2x=0\\\\2x=\pi n,nez\\\\x=\frac{\pi n }{2},nez\\\\2)Cos4x=0\\\\4x=\frac{\pi }{2}+\pi n,nez\\\\x=\frac{\pi }{8}+\frac{\pi n }{4},nez$