题目内容：

sinx-2cosx=0.求sin2x-cos2x/1+sin^2x

优质解答

sinx-2cosx=0
tanx = 2
sin2x-cos2x/1+sin^2x
=(2sinxcosx-cos²x+sin²x)/(cos²+2sin²x) [上下同除以cos²x]
=(2tanx-1+tan²x)/(1+2tan²x)
=(4-1+4)/(1+8)
=7/9