помогите решить cos x/3=-1/2

tg (2x+n/3) = - 1

sin ([/3n/6) = √3/2

sin2x=-√2/2

ь cos x/3=-1/2 x/3=+-2pi/3+2pi n x=+-2pi+6pi n

tg (2x+n/3) = - 1 2x+pi/3=-pi/4 2x=-7pi/12 x=-7pi/24

sin ([/3n/6) = √3/2 не понятен аргумент

sin2x=-√2/2 2x = (-1) ^n+1 * pi/4+pi n x = (-1) ^n+1 * pi/8+1/2pi n