已知x∈求函数-3sin^2x-4cosx 4的最大值和最小值之和

请问,会不会多了一个cosx?

f(x)=-√3sin^2x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=sin2x/2+√3cos2x/2-√3/2=sin(2x+π/3)-√3/2因为X∈[0,π/2],所以

f(x)=1+sin2x+2cos^2x=sin2x+cos2x+2=√2sin(2x+π/4)+2当x=π/8+kπ时（k为整数）f(x)取得最大值2+√2单调增区间是（kπ-3/8π,kπ+π/8

f(x)=sin²(x)+(√3)sin(x)cos(x)+2cos²(x)=3/2+√3/2sin2x+1/2cos2x=3/2+sin(2x+π/6)函数f(x)的最小正周期T

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(x)=cos(x-π/3)-sin(π/2-x)=(1/2)cosx+(√3/2)sinx-cosx=(√3/2)sinx-(1/2)cosx=sin(x-π/6）,它的最小值=-1.

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

f(x)=-√3sin²x+sinxcosx=-√3（1-cos2x）/2+（sin2x）/2=1/2sin2x+√3/2cos2x-√3/2=sin2xcos(π/3)+cos2xsin(

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

sin^2x-cosx+3=-cos^2x-cosx+4令cosx=y则原式为-y^2-y+4配方-(y+1/2)^2+17/4由x∈R则-1

已知函数f(x)=sin(x/2)+(√3)cos(x/2),x∈R；(1)求f(x)的最小正周期,并求函数f(x)在x∈[-2π,2π]上的单调增区间；（2）函数f(x)=sinx(x∈R)的图像经

y=sin²x+sinxcosx+2=(1-cos2x)/2+(sin2x)/2+2=(1/2)(sin2x-cos2x)+5/2=(1/2)*√2(sin2xcosπ/4-cos2xsin

原式=(1-cos2x)/2+(sin2x)/2+2=(sin2x-cos2x)/2+5/2=(sin（2x-45度）)*(根号2)/2+5/2所以是大于（根号2+5）/2,小于（5-根号2）/2

f（x）=sin^2x+2√3sinxcosx+3cos^2x=1+√3sin2x+2cos^2x-1+1=√3sin2x+cos2x+2=2(sin2x*√3/2+cos2x*1/2)+2=2sin

f(x)＝sin²x＋√3sinxcosx＋2cos²x＝sin²x＋√3sinxcosx＋cos²x＋cos²x＝1＋√3sinxcosx＋cos&

f(x)=√3sin(2x-π／6)+2sin^2(x-π／12)=√3sin(2x-π／6)+1-cos(2x-π／6)=2(√3/2sin(2x-π／6)-1/2cos(2x-π／6))+1=2(

1.f(x)=sin²x+2sinxcosx+3cos²x+m=2sinxcosx+2cos²x+sin²x+cos²x+m=sin(2x)+cos(

f(x)=2sinxcosx+2(cosx)^2+1=sin2x+cos2x+2=√2sin(2x+45°)+2∴当2x+45°=90°+360°k,k∈Z即x=22.5°+180°k时,f(x)取最