fx=2根号3sinxcosx+2cos平方x-1+a
f(x)=2√3sinxcosx+2(cosx)^2-1 =√3sin2x+cos2x =2(sin2xcosπ\/6+cos2xsinπ\/6)=2sin(2x+π\/6),2kπ+π\/2≤2x+π\/6≤2kπ+3π\/2,单调递减,k∈Z,函数f(x)的单调递减区间:kπ+π\/6≤x≤kπ+2π\/3,k∈Z ∴x∈[kπ+π\/6,kπ+2π\/3],k∈Z.
已知函数F(X)=2乘根号3sinxcosx+2cos^2X -1
F(X)=2乘根号3sinxcosx+2cos^2X -1=根号3sin2x+cos2x=2cos(2x-派\/3)函数最小正周期=2派\/2 =派 单调递减区间[派\/6,2派\/3]fx0=6\/5, cos(2x-派\/3)=3\/5, sin(2x-派\/3)=4\/5 , X0派\/4到派\/2 Cos2x0=(3+4根号下3)\/10 ...
fx=2•根号3sinxcosx+2cos平方x-1 化解
fx=2•根号3sinxcosx+2cos平方x-1 化解 f(x)=√3sin2x+cos2x =2[(√3\/2)sin2x+(1\/2)cos2x]=2sin(2x+π\/6)
已知函数fx=2√3sinxcosx+2cosx-1(x∈R)
1):fx=2√3sinxcosx+2cos^2 x -1 =√3sin2x+cos2x =2(√3\/2 sin2x +1\/2 cos2x)=2sin(2x+π\/6)所以最小正周期是π 2kπ-π\/2《2x+π\/6《2kπ+π\/2 kπ-π\/3《x《kπ+π\/6 单调增区间 [kπ-π\/3,kπ+π\/6] k∈Z K=0,[-π\/3,π\/6]故在【0...
已知函数f《x》=2根号3sinxcosx+2cos²x-1《x属于R》.第一题,求fx...
解:由已知条件变形得:f(x)=√3sin2x+cos2x, x∈R.f(x)=2sin(2x+π\/6).∵x∈[0,π\/2], ∴2x+π\/6∈[π\/6,π+π\/6].∵f(x)在[0,π\/6]区间为增函数,在[π\/6,π\/2]区间为减函数,∴f(x)在x=π\/6处取得最大值2;∵f(π\/2)<f(0),∴f(x)在x=π\/2处,即...
已知函数fx=2根号3sinxcosx+2cos^2X-1 求fπ\/6的值及fx的最小正周期
f(x)=√3sin2x+cos2x =2(sin2x*√3\/2+cos2x*1\/2)=2(sin2xcosπ\/6+cos2xsinπ\/6)=2sin(2x+π\/6)所以f(π\/6)=2sin(2×π\/6+π\/6)=2 T=2π\/2=π
fx=2根号3sinxcosx+cos2x+1化简
解:原式=√3sin2x+cos2x+1=2(√3\/2sin2x+1\/2cos2x+1=2cos(2x-pai\/3)+1。
已知函数fx=2根号3sinxcosx+2sin方x-1,x∈R.求函数fx的最小正周期和单 ...
f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π\/6)最小正周期T=π ,单调递增区间:2kπ-π\/2<2x-π\/6<2kπ+π\/2, :kπ-π\/6<x<kπ+π\/3 将函数y=2sin(2x-π\/6)的图像上个点的纵坐标保持不变,横坐标缩短到原来的½,得y=2sin(4x-π\/12)再把所得的...
已知函数fx=2√3sinxcosx+2cos^2x-t若方程fx=0在x属于[0,派\/2]上...
=2√3sinxcosx+2cos^2x-1+1-t =√3sin2x+cos2x+1-t =2(√3\/2sin2x+1\/2cos2x)+1-t =2sin(2x+π\/6)+1- t 由方程fx=0在x属于[0,派\/2]上有解 即2sin(2x+π\/6)+1- t=0在x属于[0,派\/2]上有解 即t=2sin(2x+π\/6)+1在x属于[0,派\/2]上有解 即t是关...
已知函数 fx=2倍的根3sinxcosx+cos的平方x-1 fx 求的最小正周期, 求最...
f(x)=2[√3sinxcosx+(cosx)^2]-1 =√3sin2x+(1+cos2x)-1 =2sin(2x+π\/6)所以,最小正周期T=2π\/2=π,最大值=2。