fx=-sin(2x 1)-x在下列区间不存在零点-搜答案-搜搜做题作业网

fx=2sinxcosx=sin2x所以最小正周期2π/2=π[-π/6,π/2]上,2x∈[-π/3,π]最小-√3/2,最大1

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

设函数fx=sin(φ-2x)(0

fx=2sin(2x+pai/6)振幅A=2最小正周期T=2pai/2=paix∈【0,pai/]2xE[0,2pai]2x+pai/6E[pai/6,2pai+pai/6]很明显,设u=2x+pai

fx=4cos²x-2+1-cos²x-4cosx=3cos²x-4cosx-1令t=cosx则-1≤t≤1即求[3t²-4t-1]的最值

0≤x≤π/20≤2x≤π-π/6≤2x-π/6≤5π/6f(x)max=f(π/3)=1f(x)min=f(0)=-1/2f(x)的值域是[-1/2,1]

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/

x∈[-π/12,π/2]2x∈[-π/6,π]2x-π/6∈[-π/3,5π/6]sin(2x-π/6)∈[-√3/2,1]2sin(2x-π/6)∈[-√3,2]值域是[-√3,2]

f(x)=(√3/2)sin2x-(1/2)[(cosx)^2-(sinx)^2]-1=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1f(x)的最大值是0,最小值是-2,

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

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

令x1=x2=0,所以f(0+0)=f(0)+f(0)+2,所以f(0)=-2令x1=x,x2=-x,所以f(x-x)=f(x)+f(-x)+2,所以f(x)+f(-x)=f(0)-2=-41.g(-

你的分析前一半是对的,一直到“那么2x的单调增区间是[-4分之π,4分之π]”.2x的单调递增区间是[-π/2,π/2],x的才是[-π/4,π/4].所以函数在x=-π/3处取得最小值为-2分之根号

(1)fx=sin(2x+φ)经过点(π/12,1)sin(π/6+φ)=1∴π/6+φ=π/2+2kπ,k∈Z∴φ=π/3+2kπ,k∈Z∵0

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

f(x)=sin(2x+π/6)-cos2x+1所以为2π/2=πf(x)=根号3/2sin2x-(cos2x)/2+1=sin(2x-π/6)+1所以最大值为2,x=π/2+2kπ-π/6=π/3+