因为F(x)=2Sinx/2的平方+cos(X—60度)=2Sinx/2的平方+cos(X—π/3)
=2×(1-cosx)/2+cosπ/3×cosx+sinπ/3×sinx
=(根3)/2倍的sinx=1/2cosx+1
=sin(x-π/6)+1
所以当sin(x-π/6)=1时,F(x)的最大值为2
F(x)的单调增区间:2kπ-π/2≤x-π/6≤2kπ+π/2即【2kπ-π/3,2kπ+2π/3】
因为F(x)=2Sinx/2的平方+cos(X—60度)=2Sinx/2的平方+cos(X—π/3)
=2×(1-cosx)/2+cosπ/3×cosx+sinπ/3×sinx
=(根3)/2倍的sinx=1/2cosx+1
=sin(x-π/6)+1
所以当sin(x-π/6)=1时,F(x)的最大值为2
F(x)的单调增区间:2kπ-π/2≤x-π/6≤2kπ+π/2即【2kπ-π/3,2kπ+2π/3】