\(P=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=2sin\left(x+\frac{\pi}{3}\right)\)
\(-1\le sin\left(x+\frac{\pi}{3}\right)\le1\Rightarrow-2\le P\le2\)
\(P_{min}=-2\) khi \(sin\left(x+\frac{\pi}{3}\right)=-1\)
\(P_{max}=2\) khi \(sin\left(x+\frac{\pi}{3}\right)=1\)
Áp dụng BĐT Bunhiakowski ta có:
\(P^2=\left(1sinx+\sqrt{3}cosx\right)^2\le\left(1+3\right)\left(sin^2x+cos^2x\right)=4\)
\(\Leftrightarrow-2\le P\le2\).
Max P = 2 khi chẳng hạn \(x=\frac{\pi}{6}\).
Min P = -2 khi chẳng hạn \(x=\frac{-\pi}{3}\).