可得sin ω(3π/4 "-" π/4)=sinπ/2ω=0,

　　∴π/2ω=kπ,k∈Z.

　　故ω的最小值是2.

答案2

10.已知函数f(x)=2sin(2x+π/3)+1.

(1)当x=4π/3时,求f(x)的值;

(2)若存在区间[a,b](a,b∈R且a

解(1)当x=4π/3时,f(x)=2sin(2×4π/3+π/3)+1=2sin(3π)+1=2sin π+1=1.

　　(2)f(x)=0⇒sin(2x+π/3)=-1/2⇒x=kπ-π/4,k∈Z或x=kπ-7/12π,k∈Z,即f(x)的零点间隔依次为π/3 和 2π/3.

　　故若y=f(x)在[a,b]上至少含有6个零点,

　　则b-a的最小值为2×2π/3+3×π/3=7π/3.

11.已知函数f(x)=2cos(ωx+π/6)(其中ω>0,x∈R)的最小正周期为10π.

(1)求ω的值;

(2)设α,β∈[0"," π/2],f(5α+5/3 π)=-6/5,f(5β"-" 5/6 π)=16/17,求cos α,sin β的值.

解(1)由已知得2π/ω=10π,∴ω=1/5.

　　(2)∵f(x)=2cos(1/5 x+π/6),

　　∴f(5α+5π/3)=2cos[1/5 (5α+5π/3)+π/6]=-2sin α,

　　f(5β"-" 5π/6)=2cos[1/5 (5β"-" 5π/6)+π/6]=2cos β.

　　又f(5α+5π/3)=-6/5,f(5β"-" 5π/6)=16/17,

　　∴sin α=3/5,cos β=8/17.

　　又∵α,β∈[0"," π/2],

　　∴cos α=4/5,sin β=15/17.

12.导学号93774035已知f(x)=Asin(2x+π/6)(A>0)的最大值为6.

(1)求A;

(2)将函数y=f(x)的图像先向左平移π/12个单位长度,再将所得图像上各点的横坐标缩短为原来的1/2倍,纵坐标不变,得到函数y=g(x)的图像.求g(x)在[0"," 5π/24]上的值域.

解(1)因为A>0,所以由题意知A=6.

　　(2)由(1)得f(x)=6sin(2x+π/6).

　　将函数y=f(x)的图像先向左平移π/12个单位长度后得到y=6sin[2(x+π/12)+π/6]=6sin(2x+π/3)的图像,

　　再将得到的图像上各点的横坐标缩短为原来的1/2倍,纵坐标不变,得到y=6sin(4x+π/3)的图像,

　　因此g(x)=6sin(4x+π/3).

　　因为x∈[0"," 5π/24],

　　所以4x+π/3∈[π/3 "," 7π/6].

　　故g(x)在[0"," 5π/24]上的值域为[-3,6].