f(x)=2sin^2(x+π/4)-√3cos2x+α在[∏/4,∏/2]上的最小值为5,求α的值

问题描述:

f(x)=2sin^2(x+π/4)-√3cos2x+α在[∏/4,∏/2]上的最小值为5,求α的值

α=4f(x)=2sin^2(x+π/4)-√3cos2x+α=2(sinxcosπ/4+cosxsinπ/4)^2-√3cos2x+α=2(√2/2sinx+√2/2cosx)^2-√3cos2x+α=2((sinx)^2/2+(cosx)^2/2+sinxcosx)-√3cos2x+α=1+2sinxcosx-√3cosx+α=1+sin2x-√3cos2x+...