Question

Tìm max, min của hàm số: y = 3sin2x + 4cos2x - 3

Answer 1

\(y=3sin2x+4cos2x-3\)

=>\(-\sqrt{3^2+4^2}-3< =y< =\sqrt{3^2+4^2}-3\)

=>-8<=y<=2

y=-8 khi \(3sin2x+4cos2x-3=-8\)

=>3sin2x+4cos2x=-5

=>\(sin2x\cdot\dfrac{3}{5}+cos2x\cdot\dfrac{4}{5}=-1\)

=>\(sin\left(2x+a\right)=-1\) với cosa=3/5

=>2x+a=-pi/2+k2pi

=>2x=-pi/2+k2pi-a

=>x=-pi/4+kpi-a/2

y=2 khi \(3sin2x+4cos2x=5\)

=>sin(2x+a)=1

=>2x+a=pi/2+k2pi

=>2x=pi/2+k2pi-a

hay \(x=\dfrac{pi}{4}+kpi-\dfrac{a}{2}\)