Решите уравнение 2cos^2x+3cosx=0

2cos^2x - 3cosx=0

2 (1-sin^2x) - 3cosx=0

2-2cos^2x-3cosx+=0

-2cos^2x-3cosx+2=0

2cos^2x+3xcos-2=0

D=9-4 (-2*2) = 9+16=25

cosx1 = (-3+5) / 4=2/4=1/2

cos2 = (-3-5) / 4=-8/4=-2

x=1/2

x = (-1) ^m*п/6+пm, m э z.