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已知F(x)=sin(2x+6/π)+2/3,x属于R

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

(1)T=2π2=π.(2)由2kπ-π2≤2x+π6≤2kπ+π2,得kπ-π3≤x≤kπ+π6,k∈Z,∴函数的单调增区间为[kπ-π3,kπ+π6](k∈Z).(3)∵x∈[0,π2],∴2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴当2x+π6=π2,即x=π6时函数有最大值1,当2x+π6=7π6时,即x=π2,函数有...

(1)∵f(x)=sin(2x+π6)+32,∴当2x+π6=2kπ-π2,k∈z,即x=kπ-π3时,函数f(x)取得最小值为-1+32=12.(2)令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,故函数的增区间为[kπ-π3,kπ+π6],k∈z.(3)把函数y=sinx的图象向左平移π6个单位...

解: f(x)=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+1+cos2x =2sin2xcosπ/6+cos2x+1 =√3sin2x+cos2x+1 =2sin(2x+π/6)+1 ⑴f(x)取得最大值3,此时2x+π/6=π/2+2kπ,即x=π/6+kπ,k∈Z 故x的取值集合为{x|x=π/6+kπ,k∈Z} ⑵由2x+π/6∈[-π/2+2kπ...

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

你是求f(x)的值域吗?

(1)f(x)=3sin(2x-π6)+2sin2(x-π12)=3sin(2x-π6)+1-cos(2x-π6)=2sin(2x?π3)+1,∴T=2π2=π由2x?π3∈[2kπ?π2,2kπ+π2]得增区间为[kπ?π12,kπ+5π12](k∈Z);(2)y=sinx右移π3得到y=sin(x-π3),纵不变,横变为原来12,得到y=sin(2x?...

(1)f(x)=sin(2x+π/6)+3/2,最小正周期为2π/2=π,单增区间为2Kπ-π/2

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