设2sin2x-cos2x sinxcosx-6sinx

f(x)=sin2x+2√3cos²x=sin2x+√3(2cos²x-1)+√3=sin2x+√3cos2x+√3=2(sin2xcos60º+cos2xsin60&#

f(x)=2(cosx)^2+sin(2x)+a=1+cos(2x)+sin(2x)+a=√2sin(2x+π/4)+a+1=√2sin[2(x+π/8)]+a+1函数最小正周期为2π/2=π2kπ-

（1）f（x）=cos（2x+π3）+sin2x=cos2xcosπ3−sin2xsinπ3+1−cos2x2＝12−32sin2x所以函数f（x）的最大值为1+32，最小正周期π．（2）由f（x）=

cos(x/2)+sin(x/2)=-√5/2[cos(x/2)+sin(x/2)]^2=(-√5/2)^2[cos(x/2)]^2+[sin(x/2)]^2+2cos(x/2)sin(x/2)=5/

采用拉格朗日记法y'=2cos2x+2*1/2(1+2x)^-1/2+0

dy/dx=[(sin2x)'(1+x²)-sin2x·(1+x²)']/(1+x²)²=[2cos2x·(1+x²)-2x·sin2x]/(1+x&

f(x)＝2cos^2x＋根号3sin2x-2=cos2x+1+√3sin2x-2=2sin(2x+π/6)-1f(π/3)=2sin(5π/6)-1=1-1=0最小正周期为2π/2=π单调增区间：2

sin^2(x)+cos^2(X)=1,原式=(3sin^2(x)+cos^2(x))/2sinxcosx,分式上下同除以cos^2(x),得原式=（3tan^2(x)+1)/2tanx,再将分子和分

f(x)=1+2sin2x+2cos^2(x)=2sin2x+(1+cos2x)+1=2sin2x+cos2x+2=√5sin(2x+φ)+2(其中cosφ=2/√5,sinφ=1/√5)

（1）f（x）=1+cos2x+sin2x+a=2sin（2x+π4）+1+a，∵ω=2，∴T=π，∴f（x）的最小正周期π；当2kπ-π2≤2x+π4≤2kπ+π2（k∈Z）时f（x）单调递增，解得

y=[2*(sinx)^2+1]/sin2x=(2-cos2x)/sin2x,x属于（0,π/2),y>0ysin2x+cos2x=2(2x+a)=2,sin(2x+a)>0sin(2x+a)=2/(

f(x)=sin2x+2cos²x+1=sin2x+2cos²x-1+2=sin2x+cos2x+2=√2(sin2xcosπ/4+cos2xsinπ/4)+2=√2sin(2x+

f(x)=sin2x+2cos²x+1=sin2x+cos2x+2=√2(√2/2sin2x+√2/2cos2x)+2=√2cos(2x-π/4)+2∴2x-π/4=2kπ时有最大值2+√2

y=(2sin²x+1)/(sin2x)=（2sin²x+sin²x+cos²x)/(2sinxcosx)=3sinx/(2cosx)+cosx/(2sinx)

ab=√3(sin2x)^2+cos2xsin2x=（√3/2）（1-cos4x）+（1/2）sin4x=（√3/2）+sin4xcos兀/3-cos4xsin兀/3=(√3/2)+sin(4x-兀/

0≤2x＜2π;-1≤sin2x≤1;sin2x=1/2；2x=π/6或5π/6;x=π/12或5π/12;

（1）f（x）=sin2x-sin（2x-π2）=1−cos2x2+cos2x＝12cos2x+12∴当cos2x=1时，函数取得最大值1；当cos2x=-1时，函数取得最小值0．（2）∵f(C2)＝

(1)f(x)=（2cosx-1）+1+sin2x+a=(cos2x+sin2x)+a+1=√2[（√2）/2sin2x+（√2）/2cos2x]+a+1=√2sin(2x+π/4）+a+1==>最小

1=sin^2x+cos^2xy=(2sin^2x+sin^2x+cos^2x)/(2sinxcosx)=(3sin^2x+cos^2x)/(2sinxcosx)y=3/2*tanx+1/2*cotx

说明：2cos^2x+sin2x/1+tanx=?有错,因为不可写成sin2x/1.我认为应该是2cos^2x+sin2x/（1+tanx）=?这样解题如下；∵(2cosx-sinx)(sinx+co