Question

2sin^2x-3cosx-3=0 укажите корни принадлежащие отрезку [пи; 3пи]

Answer 1

2sin²x-3cosx-3=0
2(1-cos²x)-3cosx-3=0
2-2cos²x-3cosx-3=0
-2cos²x-3cosx-1=0
2cos²x+3cosx+1=0
cosx=y
2y²+3y+1=0
D=3²-4*2=1
y₁=(-3+1)/4=-2/4=-0.5
y₂=(-3-1)/4=-1

cosx=-1/2
x=+-2π/3+2πk, k∈Z
[π; 3π]
x=4π/3
x=8π/3

cosx=-1
x=π+2πk, k∈Z
x=π
x=3π

Ответ 4π/3; π; 8π/3; 3π