Question

(sin^2x+1/sin^2x)(cos^2x+1/cos^2x)最小为多少

Answer 1

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