唉,只會用導(dǎo)數(shù)做出來.
令cosx=t,則原函數(shù)變?yōu)閒(t)=t³-t²-t+1
=(t+1)(t-1)²
-1≤t≤1
對t求導(dǎo),則得f(t)'=2(t-1)(t+1)+(t-1)²
=3t²-2t-1
令f(t)'=0 得t=1或者-1/3
經(jīng)演算,可知,當(dāng)t=-1/3時,原函數(shù)為最大,為32/27