Question

已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos(x+θ/2)-√3,且0≤θ≤π,求使函数f(x

Answer 1

f(x)=sin(2x+θ)+2√3[1+cos(2x+θ)]/2-√3 =sin(2x+θ)+√3cos(2x+θ) =2sin(2x+θ+π/3)是偶函数 f(-x)=f(x) 2sin(-2x+θ+π/3)=2sin(2x+θ+π/3) 所以-2x+θ+π/3=2kπ+2x+θ+π/3或-2x+θ+π/3=2kπ+π-(2x+θ+π/3) -2x+θ+π/3=2kπ+2x+θ+π/3 4x=-2kπ 不是恒等式 -2x+θ+π/3=2kπ+π-(2x+θ+π/3) 2θ=2kπ+π/3 θ=kπ+π/6 所以θ=π/6