已知tgx=2求sin^2x-3sinxcosx+4cos^2x的值

sin^2x-3sinxcosx+4cos^2x
=(sin^2x+cos^2x)+3cos^2x-3sinxcosx
=1+(3cos^2x-3sinxcosx)
=1+[(3cos^2x-3sinxcosx)/(sin^2x+cos^2x)]....分子分母同时除以cos^2x
=1+[（3-3tgx)/(tg^2x+1)]
=1+[(3-6)/(4+1)]
=2/5

sin^2x-3sinxcosx+4cos^2x
=(tg^2x-3tgx+4)*cos^2x
=(4-6+4)*cos^2x
=2cos^2x
tg^2x=4
(1-cos^2x)/cos^2x=4
cos^2x=1/5
原式=2/5

两种方法都不错
tanx=10/9
用同样的方法就可以求出
2sin^2x-sinxcosx+3cos^2x=2+(1-tanx)/(tan^2x+1)=353/81