已知函数F(x)=2根号3sinxcosx+cos²x-sin²x-1,求单调递增区间;
(1)f(x)=√3sinxcosx+cos²x =√3\/2sin2x+(cos2x+1)\/2 =√3\/2sin2x+1\/2cos2x+1\/2 =sin(2x+π\/6)+1\/2 f(x)的最小正周期为2π\/2=π 2kπ-π\/2≤2x+π\/6≤2kπ+π\/2 (k∈z)kπ-π\/3≤x≤kπ+π\/6 则f(x)的单调递增区间为[kπ-π\/3,kπ+π\/6](k...
已知函数f(x)=2√3sinxcosx+cos^2x-sin^2x-1 求周期和递增区间
f(x)=2√3sinxcosx+cos^2x-sin^2x-1 f(x)=√3sin2x+cos2x-1 =sin(2x+π\/6)-1 所以周期为:T=2π\/w=π.
已知函数f(x)=2乘以根号下3乘以sinxcosx+1-2sin平方x,x属于N,求函数f...
f(x)=2√3sinxcosx+1-2sin²x=√3sin2x+cos2x=2sin(2x+π\/6)最小正周期T=2π\/2=π 单调递增区间【kπ-2π\/3,kπ+π\/3】
已知函数f(x)=2根号3sinx.cosx+2cos2x-1求函数f(x)的单调递增区间
f(x)=根号3sin2x-cos2x+1 =2sin(2x-π\/6)+1 令2kπ-π\/2<2x-π\/6<2kπ+π\/2 kπ-π\/6<x<kπ+π\/3 所以增区间为[kπ-π\/6,kπ+π\/3]
f(x)=2根号3sinx*cosx+cos2x-1(x属于R);(1)求函数y=f(x)的单调递增区间...
f(x)=2√3sinx*cosx+cos2x-1 =√3sin2x+cos2x -1 =2sin(2x+ π\/3)-1 y=f(x)的单调递增区间 即 2kπ- π\/2≤2x+ π\/3≤ 2kπ-+π\/2 kπ- 5π\/12≤x≤ kπ-+π\/12 k∈z 2)若x属于【-5π\/12,π\/3】令k=0 x∈[-5π\/12, π\/12]递增 故f(-5π\/12)...
已知函数f(x)=2√3sinxcosx+2cos²x-1
f(x)=2√3sinxcosx+2cos²x-1 =√3sin2x+cos2x =2(√3\/2sin2x+1\/2cos2x)=2(sin2xcosπ\/6+cos2xsinπ\/6)=2sin(2x+π\/6)(1)对称轴:2x+π\/6=2kπ+π\/2 2x=2kπ+π\/3 x=kπ+π\/6;k∈Z 闭区间【0,π\/2】当x=π\/2时;函数有最小值=-2sin(π\/6)=-1...
已知函数f(x)=2根号3sinxcosx+2cos方x-1(x属于R) 求函数f(x)的单调递...
f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π\/6),则递减区间是:2kπ+π\/2≤2x+π\/6≤2kπ+3π\/2,得:kπ+π\/6≤x≤kπ+2π\/3,则减区间是:[kπ+π\/6,kπ+2π\/3],其中k∈Z
已知函数f(x)=2sin²x+2根号3sinxcosx+1
已知函数f(x)=2sin²x+2根号3sinxcosx+1,求函数的单调递增区间 函数在区间【0,п\/2】上的最值 解析:∵函数f(x)=2sin²x+2√3sinxcosx+1=-cos2x+√3sin2x+2 =2sin(2x-π\/6)+2 其单调增区间为:2kπ-π\/2<=2x-π\/6<=2kπ+π\/2==>kπ-π\/6<=x<=kπ+π...
f(x)=2根号3sinxcosx+2cos²x-1
解:f(x)=2√3sinxcosx+2cos²x-1 =√3sin(2x)+cos(2x)=2[(√3\/2)sin(2x)+(1\/2)cos(2x)]=2sin(2x+π\/6)x1∈[π\/4,π\/2]2π\/3≤2x1+π\/6≤7π\/6 cos(2x1+π\/6)<0 f(x1)=2sin(2x1+π\/6)=6\/5 sin(2x1+π\/6)=3\/5 cos(2x1+π\/6)=-√[1-...
已知函数f(x)=2根号3sinxcosx+2sin^2x-1,x 求最小正周期和单调区间.速...
f(x)=√3 sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π\/6)最小正周期T=2π\/2=π 单调增区间:2kπ-π\/2=