若(cos2x)/[sin(x-π/4)]=（-根号2）/2,则cosx+sinx=?

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(cos2x)/[sin(x-π/4)]
=[(cosx)^2-(sinx)^2]/[((根号2)/2)sinx-((根号2)/2)cosx]
=(cosx-sinx)(cosx+sinx)/[-((根号2)/2)(cosx-sinx)]
=(cosx+sinx)/[-(根号2)/2]=-(根号2)/2,
所以 sinx+cosx= 1/2.

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