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+...