f(x)=2cos^2 x+2sinx+1=2-2sin^2 x+2sinx+1=-2sin^2 x+2sinx+3=-2(sin^2 x-sinx)+3=-2(sin^2 x-sinx+1/4-1/4)+3=-2(sinx-1/2)^2+7/2x∈[-π/3,5π/6]sinx∈[-√3/2,1]因此當(dāng)sinx=1/2時(shí)有最大值f(x)=7/2,此時(shí)x=π/6或...為什么當(dāng)sinx=1/2時(shí)有最大值f(x)=7/2,此時(shí)x=π/6或5π/6sinx=-√3/2時(shí)有最大小值f(x)=(3-√3)/2,此時(shí)x=-π/3f(x)=2cos^2 x+2sinx+1=2-2sin^2 x+2sinx+1=-2sin^2 x+2sinx+3=-2(sin^2 x-sinx)+3=-2(sin^2 x-sinx+1/4-1/4)+3=-2(sinx-1/2)^2+7/2看到化簡了吧這是關(guān)于sinx的一元二次函數(shù)，因此只要前面的等于0，就得到最大值