已知4sin^2x+6sinx-cos^2x-3cosx=0,求（cos2x-sin2x）\（(1-cos2x)(1-tan2x)）的值讲述清楚答案是用什么方法解出的!

Question

已知4sin^2x+6sinx-cos^2x-3cosx=0,求（cos2x-sin2x）\（(1-cos2x)(1-tan2x)）的值
讲述清楚答案是用什么方法解出的!

Answer 1

4sin^2x+6sinx-cos^2x-3cosx=0
4sin^2x-cos^2x+6sinx-3cosx=0
(2sinx+cosx)(2sinx-cosx)+3（2sinx-cosx)=0
(2sinx-cosx)(2sinx+cosx+3)=0
∵2sinx+cosx+3≠0
∴2sinx-cosx=0
tanx=1/2
cos2x=(1-tan^2x)/(1+tan^2x)
=(1-1/4)/(1+1/4)
=3/5
（cos2x-sin2x）/[(1-cos2x)(1-tan2x)]
=（cos2x-sin2x）cos2x/[(1-cos2x)(1-tan2x)cos2x]
=（cos2x-sin2x）cos2x/[(1-cos2x)(cos2x-sin2x)]
=cos2x/(1-cos2x)
=(3/5)/(1-3/5)
=3/2