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)
當(dāng)sin(2x-π/6)=1 時(shí)取最大值,此時(shí)f(x)=4
函數(shù)的單調(diào)增區(qū)間由下式確定
-π/2+2kπ≤2x-π/6≤π/2+2kπ （k=0,±1,±2.……）
化簡得 -π/6+kπ≤x≤2π/3+kπ
即 單調(diào)增區(qū)間為[-π/6+kπ,π/3+kπ ]
加油啊![-π/6+kπ,π/3+kπ]不對f(x)=a^2+2aba^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) 當(dāng)sin(2x-π/6)=1 時(shí)取最大值,此時(shí)f(x)=4函數(shù)的單調(diào)增區(qū)間由下式確定-π/2+2kπ≤2x-π/6≤π/2+2kπ（k=0，±1，±2.……）化簡得 -π/6+kπ≤x≤π/3+kπ即 單調(diào)增區(qū)間為[-π/6+kπ,π/3+kπ]加油啊?。。?！