Question

Помогите решить первые 2 уравнения: 4.14. 4.16.

Answer 1

$1) 2cos^2x-cosx-1=0\cosx=t\2t^2-t-1=0\D=1+8=9\t_{1,2}= frac{1б3}{4} \t_1= frac{4}{4}=1, t_2= -frac{2}{4}=- frac{1}{2}\ left { {{cosx=1} atop {cosx=- frac{1}{2} }} right. textgreater left { {{x=2pi k} atop {x=б( pi-frac{pi}{3})+2pi k }} right. textgreater left { {{x=2pi k} atop {x=б frac{2pi }{3}+2pi k }} right.$

$2) 2sin^2x+7cosx+2=0\2-2cos^2x+7cosx+2=0\cosx=t\7t-2t^2+4=0\2t^2-7t-4=0\D=49+4*2*4=49+32=81\t_{1,2}= frac{7б9}{4} \t_1= frac{16}{4} = 4, t_2=- frac{2}{4}=- frac{1}{2} \ left { {{cosx=4} atop {cosx=- frac{1}{2} }} right. \\x=бfrac{2pi}{3} +2pi k$