Question

sin^2x+sin^2п/6=cos^22x+cos^2п/3
РЕШИТЕ УРАВНЕНИЕ

Answer 1

$cos^2x+sin^2frac{pi }{6}=cos^22x+cos^2frac{pi}{3}\\Formyla:; ; ; ; cos^2x=frac{1+cos2x}{2}\\ frac{1+cos2x}{2} +(frac{1}{2})^2=cos^22x+(frac{1}{2})^2\\cos^22x- frac{1}{2} cos2x-frac{1}{2} =0; |cdot 2\\2cos^22x-cos2x-1=0\\t=cos2x; ,; ; ; -1 leq t leq 1\\2t^2-t-1=0\\D=9; ,; ; t_1=-frac{1}{2}; ,; ; t_2=1\\a); ; cos2x=-frac{1}{2}; ,; ; \\2x=pm (pi -frac{pi}{3})+2pi n=pm frac{2pi}{3}+2pi n,; nin Z\\x=pm frac{pi}{3}+pi n,; nin Z$

$b); ; cos2x=1; ,; ; 2x=2pi m; ,; min Z\\x=pi m; ,; min Z$

Answer 2

А во второй строчке откуда 1+cos2x/2?