Решить уравнение (sin^2)8x=2(sin^2)4x

(sin^2)(8x)  =  2(sin^2)(4x)2*[sin(4x)]*[cos(4x)] - 2*sin(4x) = 0[sin(4x)]*[cos(4x) - 1] = 01)   sin4x = 0  4x =  πn, n∈z x1 = (πn)/4, n∈z 2)   cos4x = 1 4x = 2πk, k∈z x2 = (πk)/2, k∈z

Cos(2x-2π/3) +5sin(x-π/3)+2=0cos2(x-π/3) +5sin(x-π/3)+2=0cos²(x-π/3) - sin²(x-π/3) +5sin(x-π/3)+2=01-sin²(x-π/3) -sin²(x-π/3) +5sin(x-π/3) +2 =01 -2sin²(x-π/3) +5sin(x-π/3) +2=02sin²(x-π/3) - 5sin(x-π/3) -3 =0sin(x-π/3)=y2y² - 5y -3=0D=25 +24=49y₁=5 -7 = -1/2       4y₂ =5+7 =3        4
При у= -1/2sin(x -π/3) = -1/2x-π/3 =(-1)^(n+1) * (π/6) +πn, n∈Zx=(-1)^(n+1) * (π/6) + π/3 +πn, n∈Z
При у=3sin(x-π/3)=3Так как 3∉[-1; 1], то уравнение не имеет корней.
Ответ: (-1)^(n+1) * (π/6) +π/3 +πn, n∈Z.

1,7*10^7 км^2/3.2*10^5 км^2=53.31
Ответ примерно в 53 раза

А) х 19
б) х9
в) х (вроде так)
г) 81з