=(√3/2)*[(1+cos2x)]+(1/2)sin2x-√3/2
=√3/2cos2x+(1/2)sin2x
=sin(2x+π/3)
则最小正周期为π，
由于sinx的递增区间为[-π/2+2kπ，π/2+2kπ]
令-π/2+2kπ≤2x+π/3≤π/2+2kπ，解得-5π/12+kπ≤x≤π/12+kπ
即函数的单调增区间为[-5π/12+kπ,π/12+kπ]
2.令2x+π/3=π/2+kπ，解得x=π/12+kπ/2
而x属于(0,3π)，则x=π/12，π/12+π/2，π/12+π，π/12+3π/2，π/12+2π，π/12+5π/2
则所有相加为6*π/12+π/2+π+3π/2+2π+5π/2=8π
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已知函数f(x)=√3sinxcosx-cos²x+1\/2. 一 试求函数f(x)在区间(
回答：先把函数化简 再带入去间
帮我化简一下:函数y=√3cos²x+sinxcosx-√3\/2
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