f(x)=a^2+2ab
a^2=1,
a*b=(sinx)^2+√3sinx*cosx=1/2-1/2cos2x+√3/2sin(2x)=1/2+sin(2x-π/6)
所以 f(x)=1+2(1/2+sin(2x-π/6))=2+2sin(2x-π/6)
当sin(2x-π/6)=1 时取最大值,此时f(x)=4
函数的单调增区间由下式确定
-π/2+2kπ≤2x-π/6≤π/2+2kπ (k=0,±1,±2.……)
化简得 -π/6+kπ≤x≤2π/3+kπ
即 单调增区间为[-π/6+kπ,π/3+kπ ]
加油啊!