Question

решение задачи 2sin²x+7cоsx+2=0

Answer 1

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