От арбуза радиусом 17см отрезали верхушку(шапочку) толщиной 2см. каков радиус окружности на срезе ?

x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))

x2 = 2*pi - i*im(acos(4))

x3 = re(acos(-3)) + i*im(acos(-3))

x4 = re(acos(4)) + i*im(acos(4))

Объяснение:

x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))

x2 = 2*pi - i*im(acos(4))

x3 = re(acos(-3)) + i*im(acos(-3))

x4 = re(acos(4)) + i*im(acos(4))

x1 = 3.14159265358979 + 1.76274717403909*i

x2 = 6.28318530717959 - 2.06343706889556*i

x3 = 3.14159265358979 - 1.76274717403909*i

x4 = 2.06343706889556*i

сумма

-re(acos(-3)) + 2*pi - i*im(acos(-3)) + 2*pi - i*im(acos(4)) + i*im(acos(-3)) + re(acos(-3)) + i*im(acos(4)) + re(acos(4))

=

4*pi + re(acos(4))

произведение

(((-re(acos(-3)) + 2*pi - i*im(acos(-3)))*(2*pi - i*im(acos(4*(i*im(acos(-3)) + re(acos(-3*(i*im(acos(4)) + re(acos(4)))

=

-(2*pi - i*im(acos(4)))*(i*im(acos(-3)) + re(acos(-3)))*(i*im(acos(4)) + re(acos(4)))*(-2*pi + i*im(acos(-3)) + re(acos(-3)))

1)  2cos(2x-π/6)+√ 3 = 0      2cos(2x-π/6) = -√ 3     cos(2x-π/6)   = -√ 3/2      2x-π/6 = +- arccos(-√3/2) + 2πk , k  ∈z       2x-π/6  = +-5π/6 +2πk , k  ∈ z       2x =   +-5π/6 +2πk +  π/6  , k  ∈ z          x = +-5π/12 +  πk +  π/12, k  ∈z 2) c os(2/3x-1/3)  -1=0    c os(2/3x-1/3) =  1      2/3x-1/3 = 2πk , k  ∈ z       2/3 х = 1/3 + 2πk , k  ∈z         x = 1/2 + 3πk, k  ∈z   3)  3tg(x+5π /36)  +√   3=0        3tg(x+5π/36) = -  √   3        tg(x+5π/36) = -  √   3/3        х + 5π/36 = -π/6 +  πk , k  ∈ z         x = -π/6 +  πk - 5π/36    , k  ∈ z          х = -11π/36 +  πk , k  ∈z