设函数f(x)=cos平方x+cosx sinx（0

f(x)=cos²x+cosxsinx
=(2cos²x-1+1)/2+2sinxcosx/2
=(2cos²x-1)/2+1/2+sin2x/2
=(cos2x+sin2x)/2+1/2
=根号下2sin(2x+π/4)/2+1/2
(1) f(x)=1 则 根号下2sin(2x+π/4)/2+1/2=1
根号下2sin(2x+π/4)/2=1/2
根号下2sin(2x+π/4)=1
sin(2x+π/4)=根号下2/2
因为 x∈(0,π/2) 所以 π/4