Luonggiac-Chuong4 - CHNG IV: PHNG TRNH BAC NHA T THEO SIN...

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CHÖÔNG IV: PHÖÔNG TRÌNH BAÄC NHAÁT THEO SIN VAØ COSIN (PHÖÔNG TRÌNH COÅ ÑIEÅN) () asinu bcosu c * . a,b R\ 0 += Caùch 1 : Chia 2 veá phöông trình cho + 22 ab0 Ñaët [] ab cos vaø sin vôùi 0,2 α= α∈ π ++ c Thì * sin u cos cosu sin c sin u ⇔α = + ⇔+ α = + Caùch 2 : Neáu laø nghieäm cuûa (*) thì : uk 2 =π+ π asin bcos c b c π+ π= ⇔− = ñaët ≠π+ π 2 u tt g 2 = thì (*) thaønh : 2 2t 1 t 1t c ( )( ) 2 b c t 2at c b 0 1 vôùi b c 0 + = + Phöông trình coù nghieäm 2 'a cbcb 0 ⇔Δ= − + − ≥ 222 acb abc ⇔≥−⇔+≥ Giaûi phöông trình (1) tìm ñöôïc t. Töø u g 2 = ta tìm ñöôïc u. Baøi 87 : Tìm 26 x, 57 ππ ⎝⎠ thoûa phöông trình : cos7x 3 sin7x 2 * −= Chia hai veá cuûa (*) cho 2 ta ñöôïc : ⇔− = + = ⎛⎞ = ⎜⎟ 13 2 *c o s 7 x s i n 7 x 2 2 sin cos sin7x 66 sin 7x sin 64 2 π π = + π =+ 3 7x k2 hay 7x h2 6 4 π , ( ) k, h Z
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ππ ⇔= + = + ± 5k 2 1 1h 2 xh a y x , k , 84 7 84 7 h Do 26 x, 57 π π ⎝⎠ neân ta phaûi coù : π π <+ < < + < ± 25k 21 hay ( k, h ) 58 4 7 7 5 8 4 7 7 ⇔< + < < + < ± 2 6 2 1 1 h 2 6 hay ( k, h ) 4 7 7 4 7 7 Suy ra k = 2, = h1 , 2 5 4 53 11 2 35 Vaäy x x 84 7 84 84 7 84 11 4 59 x 84 7 84 π π =+=π = += ∨= + = π π Baøi 88 : Giaûi phöông trình ( ) 3 3sin3x 3cos9x 1 4sin 3x * −= + Ta coù : () 3 * 3sin 3x 4 sin 3x 3 cos 9x 1 −−= sin 9x 3 cos9x 1 ⇔− = 13 sin 9x cos 9x 22 1 2 = 1 sin 9x sin 32 ⎛⎞ = = ⎜⎟ 6 π π −=+ π + π ± 5 9x k2 hay 9x k2 , k 36 3 6 π π = + ± k2 7 a y x , 18 9 54 9 k Baøi 89 1 tgx sin 2x cos 2x 2 2cos x 0 * cos x −−+ = Ñieàu kieän : cos x 0 Luùc ñoù : sin x 2 * sin 2x cos2x 4 cos x 0 cos x cos x + 2 sin x sin 2x cos x cos x cos 2x 4 cos x 2 0 + = 2 sin x 1 2cos x cos x cos2x 2cos2x 0 + = = sin x cos2x cos x cos2x 0 + = ⇔=− + 0 hay sinx cosx 2 0 == = + < 2 2 cos 2x 0 nhaän do cos 2x 2 cos x 1 0 thì cos x 0 sin x cos x 2 voâ nghieäm vì 1 1 2
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() π ⇔= + ππ ⇔=+ ± ± 2x 2k 1 , k 2 k x, k 42 Baøi 90 : Giaûi phöông trình 31 8sinx * cos x sin x =+ Ñieàu kieän : sin 2x 0 Luùc ñoù (*) 2 8sin xcosx 3sinx cosx + ⇔− = + = + = + π ⎛⎞ ⎜⎟ ⎝⎠ + + π ∨= + π ⇔=+π − + ± 41 co s2xco sx 3s inx co 4 cos 2x cos x 3 sin x 3cos x 2 cos 3x cos x 3 sin x cos 3x sin x cosx 22 cos 3x cos x 3 3x x k2 3x x k2 33 k xk x , k 61 2 2 π Nhaän so vôùiñieàu kieän sin 2x 0 Caùch khaùc : (*) 2 + ( hieån nhieân cosx = 0 hay sinx = 0 khoâng laø nghieäm cuûa pt naøy ) = + 2 8(1 cos x) cos x 3 sin x cos x = + 3 8 cos x 8 cos x 3 sin x cos x = 3 6 cos x 8 cos x 3 sin x cos x = 3 13 4 cos x cos x sin x π + + π + π π ± cos 3x cos x 3 3x x k2 3x x k2 k x , k 2 2 Baøi 91 ( ) 9sin x 6cos x 3sin 2x cos2x 8 * +− += Ta coù : (*) ( ) 2 9sinx 6cosx 6sinxcosx 1 2sin x 8 + =
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() ⇔− +− ⎛⎞ ⎜⎟ ⎝⎠ 2 6 cos x 6sin x cos x 2sin x 9sin x 7 0 7 6 cos x 1 sin x 2 sin x 1 sin x 0 2 = = = + − = = += + < 222 7 1 sin x 0 hay 6 cos x 2 sin x 0 2 sin x 1 6 cos x 2sin x 7 voâ nghieäm do 6 2 7 π ⇔=+ π ∈ ± xk 2 , k 2 Baøi 92 : Giaûi phöông trình: sin 2x 2cos 2x 1 sin x
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This note was uploaded on 11/26/2011 for the course MATH 1002 taught by Professor Chuck during the Spring '11 term at University of Western States.

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Luonggiac-Chuong4 - CHNG IV: PHNG TRNH BAC NHA T THEO SIN...

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