wohennio
春芽
共回答了9个问题采纳率:88.9% 举报
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
1年前
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