Mt s dng phng trnh lng gic n ginNgi son: Trn Th Minh ChunB mn: Ton 11(NC)S tit :4 tit.3Mt S Dng Phng Trnh Lng Gic n Gin.A. Mc tiu:1. V kin thc: Gip hc sinh nm vng cch gii mt s phng trnh lng gic n gin:- Phng trnh bc nht i vi mt hm s lng gic.- Phng trnh bc hai i vi mt hm s lng gic.- Phng trnh bc nht i vi sin v cos.- dng phng trnh thun nht bc hai i vi sin v cos.- Mt s dng phng trnh c th quy v cc dng trn.2. V k nng: - Gip hc sinh nm vng cc thut ton gii thnh tho cc dng phng trnh trn.- Gip hc sinh hnh thnh mt s k nng trong vic chuyn v phng trnh lng gic n gin. Nh:+ Gp bc cao th th h bc chuyn v cc dng quen thuc.+ Thc hin php bin i v gc a v cng t loi gc cng tt.+ S dng cc cng thc bin i lng gic rt gn biu thc, chuyn v nhng dng c thut gii.- Nm vng quy tc khi t n ph cn phi c iu kin rng buc. iu kin ca cc hm lng gic chn nghim cho ng. - Bit cch biu din nghim, chn nghim trong min iu kin ca bi tonbng ng trn lng gic hay bng phng php i s.3. Thi :- T gic, tch cc hc tp.- Bit phn bit r cc khi nim c bn v vn dng trong tng trng hp c th.B. Chun b ca GV v HS:1. Gio vin: Chun b gio n chi tit, cn thn; h thng bi tp ph hp vi trnh nhn thc ca HS.2. Hc sinh: Hc bi c, nm vng cc cng thc lng gic, c trc bi mi.Phng php dy hc: Nu vn , gi m, vn pC. Tin trnh bi dy:1. Kim tra bi c.2. Ni dung bi Mi 1 Mt s dng phng trnh lng gic n ginHot ng Ca Gio Vin Hot ng Ca Hc Sinh Ni Dung-tit trcchngta hc 4 dng phng trnh lng gic c bn - y chnh l c s ta i n vicmrngraccdng phng trnh mi phc tp hn.- Chng ta hy bt u vi phngtrnhbcnht i vi mt hm s lng gic.-Mt emnhcli choc th no l phng trnh bc nht?- Trong PT0 at b + (a, b=const) nu xt t l cc hm s lng gic:sinx,cos x,tgx,cot gxth nhng dng PT c gi l PT bcnht i vi mt hm lng gic.- Qua N trn, cc em hy cho vi VDv hmbc nht i vi mt hms lng gic.- Xt VD1:3 2 3 0 tg x + .(1)Em no c th gii bi ny?- Chngta bit cch gii PT lng gic c bn, by gi nu chng ta chuyn c v dng PT c bnth bi toncgii quyt.-Phngtrnhbcnhtl PT c dng: 0 at b + vi a, b l hng s cho trc.- HS cho mt vi VD.- Nu khng c ai xung phonggiithGVtiptc I.Phngtrnhbcnht i vi mt hm s lng gic 1.nh Ngha:Phng trnhbcnht i vi mt hm s lng gic l PT c dng:0 at b + (a,b=const), trong t l mt trong cc hm s lng gic:sinx,cos x,tgx,cot gx.2 Mt s dng phng trnh lng gic n gin- Nutrongbi lmca HSkhngt KXca tgx nh li gii trn th GV cnhimvnhc HSv K:2x K +. Sau khi tm ra x nn th li xem c tha Kkhng? Ri kt lun nghim, v p n bi ny l:6 2kx +, k l nghim ca PT.- Cho HS t gii PT sau:- Gii PT: 2sin 4 0 x .- GV tng kt li kin thc.gi .- Vi gi ca GV, HS s gii:( )31 tan 2 tan3 3x _ , 6 2kx +, k .- HS gii v ch ti K sinx 1 . Nn kt lun PT v nghim.2.Phng Php Gii: chuyn v PT lng gic c bn.*Chti iukinxc nh ca hm lng gic.- Vi cch Ntng t nhphngtrnhbcnht i vi mt hmslng gic, emno Ncho c phng trnh bc hai i vi mt HS lng gic?- GV nhn xt rikt lun li ghi ln bng N.- Xt VD2: Gii PT:

22sin 5sin 3 0 x x + .(2) ( c th ku HS cho VD rixt v d hc sinh va - HS nh ngha.II.Phngtrnhbchai i vi mt hm s lng gic:1.nh Ngha:Phng trnhbc hai i vi mt HS lng gic l PT c dng:20 at bt c + + trong 3 Mt s dng phng trnh lng gic n ginnu): y l PT bc hai i visinx.-YucuHSxungphong gii( nu HS khng giic th GV hng dn tip).Gi mt HSgii PT: 22 5 3 0 t t + .(*) -SosnhhaiPTtrnthy trong PT (*) nu thayt=sinx, th trthnhPT (2). Qua nhn xt trn ai c th gii PT (2)?- Sau khi HS gii xong GV nhnxt xemHSlmc thiu K khng? Ch nhc HS khi tsinx t phi c K 1 t .- HS gii ra tmc nghim : 12t hoc3 t .- HS gii: t sinx t.K: 1 t .Khi :( ) 2 22 5 3 0 t t + . 123tt

. 3 t (loi) v1 t .Vi 12t ta c:sinx 12 26526x kx k + +

k .*asHSskhnglm bc xc nh K ca t khi tsinx t, nncuicng khng bit kt lun ra sao!t l mt trong cc hm lnggicsau:sinx,cos x,tgx,cot gx.4 Mt s dng phng trnh lng gic n gin- Xt VD3:Gii PT2cos 2 2cos 2 0 x x + .(3)- Phng trnh ban u chacdngnonhcc phng trnh hc.- 2hmslnggicli cha chung 1 gc.- Mc tiu ca ta l tm x, vychng ta phi c gng avcngmt gc, ri chuynvnhngdngPT hc.- Nh vy hoc l ta chuyn v cng mt gc l2xhoc x. Theo cc em ta nn chuyn v gc no?- Theo hng dn ca GV, HSquyt nhchuynv chunggcx, doc bi gii:( ) 3 ( )22 2cos 1 2cos 2 0 x x + ( )24cos 2cos 2 2 0 x x + + 1 2cos22cos2xx+

k .Ta c: 2cos2x 24x k t +- 1 2cos2x+ (loi) V1 212+ < .Kt lun:NghimcaPT: 24x k t +,k .- HS thy vic t cos x t 2. Phng php gii:- ttbng cc hm lng gicsinx,cos x,tgx,cot gx(tytheobi m c cch t cho ph hp).* Ch : xc nh K ca t. - VD: Nu tsinx t(cos x t ) thK: 1 t .- Chuyn v gii phng trnh bc hai i vi n t.- Tm ra t nh kim tra K loi trmt sgi tr ca t.-Giiphngtrnhtheox vi cc gi tr t tng ng.5 Mt s dng phng trnh lng gic n gin- Xt VD4: Gii PT:( )3cot 3 1 0 tgx gx + + (4);2 2x _ ,.- gii bi trn GV gi cho HS:- Khi PT c cha tgx,cot gx ta nn ch t K ntn?- Cho c bit giatgx,cot gxcmiquanhnh th no?-Thdavoiu chuyn (b) v dng quen thuc, c khng?GVch quan st tng bc gii ca HS. Ch nhc HS t K.- Sau khi HS lm xong, GV nhn xt phn kt lun l khng cn thit, c xem nhlPTbchai i vi bin cos xm gii bnh thng.- 1cottgxgx2x k +, i vi tgxx k i vi cot gx.- Da vo gi ca GV hs gii nh sau:(4)( )33 1 0 tgx tgxtgx+ + ( )23 1 3 0 tg x tgx + + .t tgx t .Gii tm c3 t v 1 t .Khi : tan 33x x k +tan 14x x k +,k Kt lun:NghimcaPT l:3x k +v4x k + k .6 Mt s dng phng trnh lng gic n ginnghim, chrachsaica HS: nu ly 1 k th em xem 2 nghim c thuc khong;2 2 _ , khng?- Kt hp vi k ca bi cho ;2 2x _ ,th ta cn phi la chn k sao cho ph hp.- C s hng dn cc em xc nh k bng 2 cch: th nht l da vo ng trn lng gic v th 2 l bng pp i s. Cch 1:da vo ng trn lng gic.-GVv ngtrnlng gicri bohcsinhxc nh v tr3 ; 4; 2;2.-Ch trnngtrncho HS bit h nghim3k +; 4k +.- Ch cho HS bit TX ca bi;2 2x _ ,th s l cung no trn ng trn. - Sau bo HS quan st xem h nghim3k +; 4k + th c nhngnghimno - Vi 1 k HS tnh ra nghim v nhn thy nghim khng thuc;2 2 _ ,.- HS nhn bit c ch c nghim3v4l thuc 7 Mt s dng phng trnh lng gic n ginnm trong cung;2 2 _ ,.Cch2:Phngphpi s.- V nghim;2 2x _ , nn ta c:

2 3 2k < +