f(x)=cos^2x+√3sinxcosx-1/2
=1/2*(2cos^2x-1)+√3/2*sin2x
=1/2cos2x+√3/2*sin2x
=sin(2x+π/6)
x∈[0,π/2],
因为要取最大值,所以
2x+π/6=π/2
2x=π/3
x=π/6
从而
当x=π/6时,f(x)有最大值=1.
f(x)=cos^2x+√3sinxcosx-1/2
=1/2*(2cos^2x-1)+√3/2*sin2x
=1/2cos2x+√3/2*sin2x
=sin(2x+π/6)
x∈[0,π/2],
因为要取最大值,所以
2x+π/6=π/2
2x=π/3
x=π/6
从而
当x=π/6时,f(x)有最大值=1.