Tryg Tommy:

	π		2π		π
sin2(x+		)=sin(x−		)*sin(x+		)
	3		3		3

Korzystając z

	x+y		x−y
sinx−siny=2cos		sin
	2		2

Za nic mi nie może wyjść poprawny wynik

Basia: sin²(x+^π₃)−sin(x−^2π₃*sin(x+^π₃=0 sin(x+^π₃)*[ sin(x+^π₃−sin(x−^2π₃)]=0 1. sin(x+^π₃)=0 x+^π₃=kπ

	3kπ−π		(3k−1)π
x=kπ−π3=		=
	3		3

2. sin(x+^π₃)−sin(x−^2π₃)=0

x+π3+x−2π3		2x−π3
	=		= x−π6
2		2

x+π3−x+2π3		3π3
	=		= π2
2		2

2cos(x−^π₆)*sin^π₂=0 2cos(x−^π₆)*1=0 cos(x−^π₆)=0 x − ^π₆=(2k+1)^π₂

	3(2k+1)π+π		(6k+4)π		(3k+2)π
x=(2k+1)π2+π6=		=		=
	6		6		3