Cos(2x-兀 3)的单调递减区间-搜答案-百度作业网

因为单调递减,所以2kπ《π/3-x/2《2kπ+π2kπ-π/3《-x/2《2kπ+2π/34kπ-2π/3《-x《4kπ+4π/3所以4kπ-4π/3《-x《4kπ+2π/3周期T=2π/w,已知

y有最大值x取值集合:{兀/3+k兀,k是整数}单调递减区间:(k兀-2兀/3,k兀-兀/6),k是整数

求f(x)=2sin^2(π/4-x)-2倍根号3cos^2x+根号3的最小正周期和单调递减区间解析：∵f(x)=2sin^2(π/4-x)-2√3cos^2x+√3=(cosx-sinx)^2-2√

y=sin(x+π/2)cos(x+π/6)=cosx*cos(x+π/6)=cosxcosx1/2根号3+1/2cosxsinx=1/2根号3cos^2x+1/4sin2x=1/2根号3*1/2(1

y=cos(π/3-x/2)就是y=cos(x/2-π/3)就是求cos(x/2-π/3)的递减区间x/2-π/3属于[2kπ,π+2kπ]所以x属于[2π/3+2kπ,8π/3+2kπ]

y=1/2cos^x+√3/2sinxcosx+1=(1+cos2x)/4+√3/4sin2x+1=(√3/4sin2x+1/4cos2x)+5/4=1/2sin(2x+π/6)+5/4所以2kπ-π

解析求导y'=2x-12x-1>=0函数递增2x>=1x>=1/2所以函数在[1/2+无穷）递增在（-无穷1/2]单调递减

余弦倍角公式化简y=-cos2x所以单减区间为-Pai+2kPai即-Pai/2+KPai

求函数y=cos²x-cosx+2的单调递减区间与单调递增区间解析：∵f(x)=(cosx)^2-cosx+2令f’(x)=-2(cosx)sinx+sinx=sinx(1-2cosx)=0

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≤

原式=lgcos2x=根据复合函数同增异减求cos2x在（0,1】的减函数区间2kπ

【导入】此类题目应该用整体法.令-2x+π/3=T∵x∈[0,π]∴T∈[-5π/3,π/3]∴y=cosTT∈[-5π/3,π/3]根据y=cosx的函数性质y=cosT在T∈[-5π/3,-π]∪

减区间通式2x+π/3∈[2kπ-π,2kπ]k∈Zx∈[kπ-2/3π,kπ-π/6]k∈Z∵x∈[0,π]令k=0则x∈[0,5/6π]

y=cos²x-cosx+2=(cosx-1/2)^2+2-1/4当1/2

cos(-x/3+pai/4)=cos(x/3-/4)>02kπ-π/2

y=3cos(2x-π/6)+2cos的减区间是(2kπ,2kπ+π)所以2kπ

-cos2x递减则cos2x递增所以就是2kπ-π

y=cos(π/4-2x）=cos(2x-π/4）∵y=cosx在[2kπ,(2k+1)π]上为减函数∴2kπ≤2x-π/4≤(2k+1)π∴π/8+kπ≤x≤5π/8+kπ∴y=cos(π/4-2x

∵对于函数y=cos(2x-π3)的单调减区间为2kπ≤2x-π3≤2kπ+π即kπ+π6≤x≤kπ+2π3故函数f（x）的单调减区间为[kπ+π6，kπ+2π3]（k∈Z）故答案为：[kπ+π6，k