решить уравнение sin²x-3sinxcosx+2cos²x=0

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.

Всего 15 мячиков
4/15=0,26