【正文】
xxxxyxxxyxxyxxxy)三。求都有若axfxaxxxf,2|)(|]2,0[,2s i n3c o s2)(.22???????02c oss i n2.32 有解的方程為何值時(shí),關(guān)于實(shí)數(shù)??? mxxxm的最值值,并求此時(shí)的)求使(表達(dá)式)寫出(最小值記為)(21)(2)(1)(s i n2c os221)(.42xfaagagagxxaaxf?????四:求最小正周期 xyxyxyxyxxyxyxxyxyxyxxy2s i n)10|2t a n|)9 |2|s i n8) |s i n|)7 t a n1t a n16) |)52c o s (|)5 s i n12t a n4) 33t a n)3 )132c o s (2)22c o s42s i n3)1222??????????????????五:判斷函數(shù)奇偶性 xxyxxxxxxxfxxyxxxyxxxxyxys i n1s i n1)6 0,220,22{)()5t a nt a n14) 11)1()3c o