发布时间：2019-07-29 18:47:20
求x，以及正负号
cos(2α)-sin(2α)=√x*sin(2α±π/4)
当sin(2α±π/4)≠0时
√x=√2*cos(2α+π/4)/sin(2α±π/4)
x=2cos²(2α+π/4)/sin²(2α±π/4)
=2[1+cos(4α+π/2)]/[1-cos(4α±π/2)]
=2[1-sin(4α)]/[1±sin(4α)]
x₁=2[1-sin(4α)]/[1+sin(4α)]
x₂=2