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幼苗
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设t=(sinx)^2,则
原式=（t+1/t)[1-t+1/(1-t)]
=t-t^2+t/(1-t)+(1-t)/t+1/(t-t^2)
=t-t^2-2+1/(1-t)+1/t+1/(t-t^2),记为f(t),
f'(t)=1-2t+1/(1-t)^2-1/t^2-(1-2t)/(t-t^2)^2,
令f'(t)=0,得（1-2t)(t-t^2)^2+t^2-(1-t)^2-1+2t=0,
整理得2t^5-5t^4+4t^3-t^2-4t+2=0,0
1年前
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