tan(π+α)=3
tana=3
2cos(π-α)-3sin(π+α)/4cos(-α)+sin(2π-α)
=(-2cosa+3sinα)/(4cosα-sinα)
=(-2+3tanα)/(4-tanα)
=7
已知tan（π+α）=3,求《2cos（π-α）-3（π+α）》÷《4cos（-α）+sin（2π-α）》的值
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