f(x)=2sin^2-2根号3*sinX*cosX+1的周期怎么求如题,

已知函数f(x)=2根号3sin平方x-sin（2x-π/3) 函数f(x)=负根号3sin ^2x+sin x cos x f(a/2)=1/4-根号3/2求sin a值 已知F(X)=根号3COS^2 X+SIN XCOS X-2SIN X*SIN(X-π/6),求F(X)的最大值 求函数f(x)=2sin(π-x)sin(π/2-x)+2根号3sin^2x-根号3的单调递减区间 f(x)=2sin^2(π/4+x)+根号3(sin^2x+cos^2x)f(x)=2sin^2(π/4+x)+根号3(sin^2x-cos^2x) 符号打错了 f(x)=sin(2x-45)-2根号2*sin方x的周期 f(x)=2sinπ/4cosπ/4- 2根号3sin^2π/4+根号3 f(x)=2sinπ/4cosπ/4-2根号3sin^2π/4+根号3 f(x)=2sin^2x+根号3sin2x+α,且|f(x)| f(x)=1+sin(2x)-根号3 cos2x,|f(x)-m| f(x)=1+sin(2x)-根号3 cos2x |f(x)-m| f(x)=根号3sin^2x+sinxcosx-根号3/2 f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间 傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si 已知函数f(x)=根号3sin(2x+fai)-cos(2x+fai)(0 函数f(x)=sin(2x+q)+根号3cos(2x+q)(0 已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0 化简F(X)=2COSXCOSX(π/6-X)-根号3SIN^2X+SINXCOSX