已知函数f(x)=2cos(π/3-2x) .(1)若F(x)=1,x∈[-6/π,π/4],求x的值(2)求f(x)的单调增区间
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已知函数f(x)=2cos(π/3-2x) .(1)若F(x)=1,x∈[-6/π,π/4],求x的值(2)求f(x)的单调增区间
已知函数f(x)=2cos(π/3-2x) .(1)若F(x)=1,x∈[-6/π,π/4],求x的值(2)求f(x)的单调增区间
已知函数f(x)=2cos(π/3-2x) .(1)若F(x)=1,x∈[-6/π,π/4],求x的值(2)求f(x)的单调增区间
(1)∵-π/6≤x≤π/4 ∴-π/3≤2x≤π/2,-2π/3≤2x-π/3≤π/6
又∵2cos(π/3-2x)=1,即:cos(2x-π/3)=1/2
∴2x-π/3=-π/3,解得x=0
(2) 单调增区间:2kπ-π≤2x-π/3≤2kπ,
即:2kπ-2π/3≤2x≤2kπ+π/3
∴kπ-π/3≤x≤kπ+π/6,k为整数
即:f(x)的单调增区间为〔kπ-π/3,kπ+π/6〕
(1) 2cos(π/3-2x)=1,即:cos(2x-π/3)=1/2
因为:-π/6≤x≤π/4,所以-π/3≤2x≤π/2,即-2π/3≤2x-π/3≤π/6
所以2x-π/3=-π/3,即x=0
(2) 单调增区间:2kπ-π≤2x-π/3≤2kπ,即:2kπ-2π/3≤2x≤2kπ+π/3
即:kπ-π/3≤x≤kπ+π/6,k为整数
单调减区间...
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(1) 2cos(π/3-2x)=1,即:cos(2x-π/3)=1/2
因为:-π/6≤x≤π/4,所以-π/3≤2x≤π/2,即-2π/3≤2x-π/3≤π/6
所以2x-π/3=-π/3,即x=0
(2) 单调增区间:2kπ-π≤2x-π/3≤2kπ,即:2kπ-2π/3≤2x≤2kπ+π/3
即:kπ-π/3≤x≤kπ+π/6,k为整数
单调减区间:2kπ≤2x-π/3≤2kπ+π,即:2kπ+π/3≤2x≤2kπ+4π/3
即:kπ+π/6≤x≤kπ+2π/3,k为整数
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1)2cos(π/3-2x)=1
得:cos(π/3-2x)=1/2
π/3-2x=2kπ+π/3, 或π/3-2x=2kπ-π/3
x=-kπ或x=-kπ+π/3
在[-π/6, π/4]区间内,只有解x=0
2)f(x)写为 f(x)=2cos(2x-π/3)
单调增区间为: (2k-1)π=<2x-π/3<=2kπ
即kπ-π/3=