4sin^2x+9cosx-6=0 6sin^2+5sinx-11=0

1)4-4cos²x+9cosx-6=04cos²x-9cosx+2=0cosx=a4a²-9a+2=0D=81-32=49a1=(9-7)/8=1/4⇒cosx=1/4⇒x=+-arccos1/4+2πn,n∈Za2=(9+7)/8=2⇒cosx=2>1 нет решения2)sinx=a6a²+5a-1=0D=25+24=49a1=(-5-7)/12=-1⇒sinx=-1⇒x=-π/2+2πn,n∈Za2=(-5+7)/12=1/6⇒sinx=1/6⇒x=(-1)^k*arcsin1/6+πk,k∈Z