Question

Giải giúp me theo pp hạ bậc

1/ (3m - 2) sin^2x - (5m - 2) sin2x + 3(2m + 1) cos^2x = 0 ( vô nghiệm)

Answer 1

=>\(\left(3m-2\right)\cdot\dfrac{1-cos2x}{2}-\left(5m-2\right)\cdot sin2x+\left(6m+3\right)\cdot\dfrac{1+cos2x}{2}=0\)

=>(3m-2)*1/2-1/2*cos2x(3m-2)-(5m-2)*sin2x+(6m+3)*1/2+(6m+3)*1/2*cos2x=0

=>cos2x[-3/2m+1+3m+3/2]+sin2x(-5m+2)=-1/2(3m-2)-1/2*(6m+3)=-1/2(3m-2+6m+3)=-1/2(9m+1)

=>cos2x(3/2m+5/2)+sin2x(-5m+2)=-1/2(9m+1)

=>cos2x(3m+5)+sin2x(-10m+4)=-9m-1

Để phương trình có nghiệm thì (3m+5)^2+(-10m+4)^2>=(-9m-1)^2

=>9m^2+30m+25+100m^2-80m+16>=81m^2+18m+1

=>109m^2-50m+41-81m^2-18m-1>=0

=>28m^2-68m+40>=0

=>m<=1 hoặc m>=10/7