Решить тригонометрическое уравнение:

2sin²x-3cosx-3=0, [π; 3π]

1) 2 (1-cos²x) - 3cosx-3=0

2-2cos²x-3cosx-3=0

-2cos²x-3cosx-1=0

2cos²x+3cosx+1=+

cosx=t, |t|≤1

2t²+3t+1=0

t=-1 или t=-1/2

cosx=-1 cosx=-1/2

x=π+2πn x=-π/2 + 2πn

2) π≤π+2πn≤3π π≤-π/2+2πn≤3π

0≤2πn≤2π 3π/2≤2πn≤7π/2

0≤n≤1 3/4≤n≤7/4

n=1 n=1

x=π+2π*1=π+2π=3π x=-π/2+2π*1=3π/2