Question

已知tgx= -2,求(sin*2x+3cos*2x)/(3sin*2x-cos*2x)

Answer 1

(sin*2x+3cos*2x)/(3sin*2x-cos*2x)(上下同除以cos^2x)

=(tg^2x+3)/(3tg^2x-1)

=(4+3)/(3*4-1)

=7/11