Question

tìm gtln

\(\sqrt{1+2cos^2x}\) + \(\sqrt{1+3sin^2x}\) =y

Answer 1

\(y=\sqrt{1+2cos^2x}+\sqrt{1+3\left(1-cos^2x\right)}=\sqrt{1+2cos^2x}+\sqrt{4-3cos^2x}\)

\(y=\sqrt{2}.\sqrt{\dfrac{1}{2}+cos^2x}+\sqrt{3}.\sqrt{\dfrac{4}{3}-cos^2x}\)

\(y\le\sqrt{\left(2+3\right)\left(\dfrac{1}{2}+cos^2x+\dfrac{4}{3}-cos^2x\right)}=\dfrac{\sqrt{330}}{6}\)

\(y_{max}=\dfrac{\sqrt{330}}{6}\) khi \(cos^2x=\dfrac{7}{30}\)