作业帮已知tanx=-1/3,求（2+5cos2x）/(3+4sin2x)
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已知tanx=-1/3,求（2+5cos2x）/(3+4sin2x)
已知tanx=-1/3,求（2+5cos2x）/(3+4sin2x)
已知tanx=-1/3,求（2+5cos2x）/(3+4sin2x)
tan(x) = -1/3
tan(x)^2 = sin(x)^2/cos(x)^2 = [1-cos(x)^2]/cos(x)^2 = 1/9
cos(x)^2 = 9/10
cos(2x)=2*cos(x)^2 -1 =4/5
sin(2x) = 2sin(x)cos(x) = 2*sin(x)/cos(x)*cos(x)^2 =2tan(x)cos(x)^2 = 2*(-1/3)(9/10) = -3/5
（2+5cos2x）/(3+4sin2x) = 10