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共回答了22个问题采纳率:95.5% 举报
∵2sin²α-cos²α+sinαcosα-6sinα+3cosα=0
==>(sinα+cosα)(2sinα-cosα)-3(2sinα-cosα)=0
==>(2sinα-cosα)(sinα+cosα-3)=0
==>2sinα-cosα=0 (sinα+cosα-3≠0)
==>cosα=2sinα
==>4sin²α+sin²α=1 (由sin²α+cos²α=1得)
==>sin²α=1/5,tanα=1/2
∴(2cos²α+sinαcosα)/(1+tanα)
=(2*4sin²α+2sinα*sinα)/(1+tanα)
=10sin²α/(1+tanα)
=10*(1/5)/(1+1/2)
=4/3.
1年前
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