Решение задачи 2sin²x+7cоsx+2=0
2sin²x+7cosx+2=0 2(1-cos²x)+7cosx+2=0 -2cos²x+7cosx+4=0 2cos²x-7cosx-4=0 D=49+32=81 √81=9 cosx=1/4[7-9]=-1/2 cosx=1/4[7+9]=4>1 cosx=-1/2 x=+-2π/3+2πk k∈Z