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Luonggiac-Chuong2

# Luonggiac-Chuong2 - C hng 2 PHNG TRNH L N G GIA C C BA N u...

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Chöông 2: PHÖÔNG TRÌNH LÖÔÏNG GIAÙC CÔ BAÛN =+ π =⇔ =π− + π uvk 2 sin u sin v uv k 2 cosu cos v u v k2 = ± + π π ≠+ π uk tgu tgv 2 ' ( ) k,k ' Z cot gu cot gv ' ≠π Ñaëc bieät : si n u 0 u k =⇔=π π = ⇔=+π co su 0 u k 2 ( sin u 1 u k2 k Z 2 π =⇔= + π ∈ ) 1 u k2 = ⇔= π () kZ sin u 1 u k2 2 π =− ⇔ =− + π 1 u k2 =−⇔ =π + π Chuù yù : sin u 0 1 ≠⇔ ± 0 sin u 1 ± Baøi 28 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2002) [ ] x0 , 1 4 nghieäm ñuùng phöông trình Tìm ( ) cos3x 4 cos 2x 3cos x 4 0 * −+ = Ta coù (*) : ( ) 32 4 cos x 4 2cos x 1 4 0 −− + = 4cos x 8cos x 0 = ( ) 2 4cos x cosx 2 0 = ( ) == cosx 0hay cosx 2 loa ïi vìcosx 1 xk k 2 π =+ π∈ Z Ta coù : [] , 1 4 0 k 1 2 4 π ∈⇔ + π k1 4 22 ππ −≤π ≤ − 11 4 1 0, 5 k 3, 9 −= ≤− π Maø k neân Z { } k . Do ñoù : 0,1,2,3 357 x ,,, 2222 π πππ ⎩⎭ Baøi 29 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2004) Giaûi phöông trình : ( ) ( ) 2cos x 1 2sin x cos x sin 2x sin x * =

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Ta coù (*) () ( ) ( ) −+ = 2cos x 1 2sin x cos x sin x 2cos x 1 ( ) 1 cos x sin x 0 +− ⎡⎤ ⎣⎦ = ) ( 2cosx 1 sinx cosx 0 += 1 cos x sin x cos x 2 =∨ = cos x cos tgx 1 tg 34 ππ ⎛⎞ = = ⎜⎟ ⎝⎠ =± + π∨ =− + π xk 2 , k Z Baøi 30 : Giaûi phöông trình + ++= cos x cos2x cos3x cos4x 0(*) ( ) cos x cos3x 0 +++= 5x 3x 5x x 2cos .cos 0 22 2 2 5x 3x x cos cos 0 2 5x x 4 cos cos x cos 0 = 5x x cos 0 cos x 0 cos 0 = ∨= = π =+ π ∨=+ π π 5x x kx k k 2 2 2 π π π + π 2k xx k x 2 , 55 2 k Z Baøi 31: ( ) 2 2 sin x sin 3x cos 2x cos 4x * +=+ 1111 1 1 cos6x 1 1 cos8x 2222 −= ++ + = + 2cos4x cos2x 2cos6x cos2x ( ) 2cos2x cos6x 0 4 cos2x cos5x cos x 0 = 0 cos5x 0 cos x 0 = π π + π π∈ ± 2x k 5x k x k ,k 2 π π π + ∨=+π k kk x 42 1 05 2 ± ,k Baøi 32 : Cho phöông trình π x7 sin x.cos 4x sin 2x 4 sin * 42 2 Tìm caùc nghieäm cuûa phöông trình thoûa: < x1 3
Ta coù : (*) () 17 sin x.cos4x 1 cos4x 2 1 cos x 22 ⎡π ⎛⎞ 2 −= ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ −+ = −− 11 3 sin x cos 4x cos 4x 2sin x 2 1 sin x cos4x cos 4x 1 2sin x 0 2 ++ + = += cos4x sin x 2 sin x 0 1 cos 4x 2 sin x 0 2 + 2 loaïi 1 sin x sin 26 =− π =− = π = −+ π π = xk 6 7 x2 6 2 h Ta coù : < x1 3 3x13 −< − < 2x4 << Vaäy : 2k 2 6 π −<− + π< 4 k 4 66 ππ −< π <+ 21 k 12 12 −<<+ Do k neân k = 0. Vaäy Z x 6 π = − π −< + π< 7 2h 2 6 4 π π < π< − ⇔− − < < − π π 77 1 7 2 2 4 h 1 2 7 1 2 h = 0 π = 7 x 6 .Toùm laïi == 7 xh a y x Caùch khaùc : −π =− ⇔ = − + π ± k 1 sin x x ( 1) k , k −π −<− +π< ⇔ <− + < π π kk 2( 1 ) ( 1 ) k 4 k=0 vaø k = 1. Töông öùng vôùi 7 a y x Baøi 33 : Giaûi phöông trình ( ) 333 sin x cos 3x cos x sin 3x sin 4x * ( ) 33 3 3 3 sin x 4 cos x 3cos x cos x 3sin x 4 sin x sin 4x = 3 3 3 4 sin x cos x 3sin x cos x 3sin x cos x 4 sin x cos x sin 4x −+− = 3 3sin x cos x cos x sin x

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Luonggiac-Chuong2 - C hng 2 PHNG TRNH L N G GIA C C BA N u...

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