已知函数f(x)人sin2x十√3sinXsin(x十2分之兀)求最小正周期
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(Ⅰ)因为f(x)=sin2x-(1-cos2x)=2sin(2x+π4)-1,所以函数f(x)的最小正周期为T=2π2=π(Ⅱ)由(Ⅰ)知,当2x+π4=2kπ−π2,即x=kπ−π8(k∈Z)时,
f(x)=sin2x-cos2x+1=√2*(√2/2*sin2x-√2/2*cos2x)+1=√2sin(2x-π/4)+1最小正周期为:T=2π/2=π∵-1≤sin(2x-π/4)≤1∴1-√2
A=2,T=π∴ω=2∴f(x)=2sin(2x+φ﹚过﹙π/6,2﹚∴2sin(π/3+φ﹚=2sin(π/3+φ﹚=1π/3+φ=2kπ+π/2φ=2kπ+π/6∴φ=π/6∴f(x)=2sin(
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
分式有意义,cosx≠0f(x)=[sin(2x)+cos(2x)+1]/(2cosx)=(2sinxcosx+cos²x-sin²x+cos²x+sin²x)
函数f(x)=(sinx一cosx)sin2x/sinxsinx≠0,所以x≠kπ,k∈Z.函数定义域是{x|x≠kπ,k∈Z}.f(x)=(sinx一cosx)sin2x/sinx=(sinx一co
f(x)=sin2x+2cos²x-1+1=sin2x+cos2x+1=√2(√2/2*sin2x+√2/2cos2x)+1=√2(sin2xzosπ/4+cos2xsinπ/4)+1=√2
∵f(x)=sin2x=2sin²x-1∴令u(x)=sinx∴f(u)=2x²-1f(u)在(-∞,+∞)上是递增函数u(x)在(2kπ-½π,2kπ+&frac1
解.(1)函数f(x)=2sin2x+sin2x-1=sin2x-cos2x=2sin(2x-π4)…(3分)所以f(x)的最小正周期是π,…(4分)当2x-π4=2kπ+π2,k∈Z,即x=kπ+3
(Ⅰ)∵f(x)=3sin2x-2sin2x=3sin2x+cos2x-1=2sin(2x+π6)-1故函数f(x)的最大值等于2-1=1(Ⅱ)由f(x)=0得23sinxcosx=2sin2x,于是
f(x)=sin2x+cos2x-1=√2sin(2x+π/4)-1.1、最小正周期是π,最大值时2x+π/4=2kπ+π/2,即x=kπ+π/4,k是整数.再问:已知函数f(x)=2sin(∏-X)
(1)f(x)=2sinxcosx+2cos2x2cosx=sinx+cosx(cosx≠0),…(4分)由题意可得f(x)=2sin(x+π4)=0,故x+π4=kπ,即 x=kπ−π4(
已知函数f(x)=根号3sin2x+cos2x+21求f(x)的最大值及f(x)取得最大值时自变量x集合f(x)=根号3sin2x+cos2x+2=2[(根号3/2)sin2x+(1/2)cos2x]
f(x)=√3sin2x+cos2x=2(cosπ/6sin2x+sinπ/6cos2x)=2sin(2x+π/6)单调递增区间为:2kπ-π/2≤2x+π/6≤2kπ+π/2解得:kπ-π/3≤x≤
(1)f(x)=3sin2x+cos2x=2(sin2xcosπ6+cos2xsinπ6)=2sin(2x+π6)∴T=2π2=π,当2x+π6=2kπ+π2,k∈Z,即x=π6+kπ,k∈Z时,函数
f(x)=√2sin(2x+π/4)-1(1)最小正周期π;最大值√2(2)2kπ-π/2≤2x+π/4≤2kπ+π/22kπ-3π/4≤2x≤2kπ+π/42kπ-3π/8≤x≤2kπ+π/8
(1)由已知f(x)=3sin2x+2cos2x+3=3sin2x+cos2x+4=2sin(2x+π6)+4.(3分)当x∈(0,π2)时,2x+π6∈(π6,7π6),sin(2x+π6)∈(−1
sqrt表示根号下;首先将原函数化成正弦型函数f(x)=sin2x+cos2x+2=(sqrt2)*(sin2x*(sqrt2)/2+cos2x*(sqrt2)/2)+2=(sqrt2)*sin(2x
(1)∵f(x)=sin2x+3cos2x=2sin(2x+π3),故[f(x)]max=2,[f(x)]min=2.(2)函数的最小正周期为T=2π2=π.(3)令2kπ-π2≤2x+π3≤2kπ+
f(x)=sin2x-3(2cos2x-1)=sin2x-3cos2x=2(12sin2x-32cos2x)=2sin(2x-π3),∵ω=2,∴函数f(x)的最小正周期是2π2=π,当-π2+2kπ