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

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