f(x)=sin²ωx+√3cosωxXcos(π/2-ωx)(ω>0),且函数y=f(x)的图像相邻两条对称轴之间的距离为π/2.

问题描述:

f(x)=sin²ωx+√3cosωxXcos(π/2-ωx)(ω>0),且函数y=f(x)的图像相邻两条对称轴之间的距离为π/2.
(1)求ω的值及f(x)的单调递增区间
(2)在△ABC中,a,b,c分别是角A,B,C的对边,若a=√3,b=√2,f(A)=3/2,求角C

(1)化简得y=sin(2ωx-π/6)+1/2,由y=f(x)的图像相邻两条对称轴之间的距离为π/2.
则T=π,ω=1,单调递增,2kπ-π/2