Ответ:
sin^2x+sin2x=1
sin^2x+2sinxcosx-1=0
sin^2x+2sinxcosx-sin^2x-cos^2x=0
2sinxcosx-cos^2x=0
cosx(2sinx-cosx)=0
1)cosx=0,x=π/2+πk,k-Z
2)2sinx-cosx=0
2sinx=cosx |:cosx
2tgx=1
tgx=1/2
x=arctg1/2+πn,n-Z
956) -4x-1<0⇔4x+1>0,
957)-2x-9<0⇔2x+9>0,
958)-10x-8<0⇔10x+8>0,
959)-10x-1<0⇔10x+1>0,
y'=3x²+24x+36=0
x=-6 п.к
x=-2
y(-2)=-8+48-72+3=-29
y(-5)=-125+300-180+3=-2
y(-0.5)=-0,125+3-18+3=-12,125