Chöông 2: PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CÔ BAÛ N

⎡ u = v + k2π sin u = sin v ⇔ ⎢ ⎣ u = π − v + k2π cos u = cos v ⇔ u = ± v + k2π π ⎧ ⎪u ≠ + kπ tgu = tgv ⇔ ⎨ 2 ⎪⎩u = v + k ' π ⎧u ≠ kπ cot gu = cot gv ⇔ ⎨ ⎩u = v + k ' π Ñaë c bieä t : sin u = 0 ⇔ u = kπ

π + k2π ( k ∈ Z) 2 π sin u = −1 ⇔ u = − + k2π 2 Chuù yù : sin u ≠ 0 ⇔ cos u ≠ ±1 cos u ≠ 0 ⇔ sin u ≠ ±1 sin u = 1 ⇔ u =

( k, k ' ∈ Z )

cos u = 0 ⇔ u =

π + kπ 2

cos u = 1 ⇔ u = k2π ( k ∈ Z ) cos u = −1 ⇔ u = π + k2π

Baø i 28 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i D, naê m 2002) Tìm x ∈ [ 0,14 ] nghieä m ñuù ng phöông trình cos 3x − 4 cos 2x + 3 cos x − 4 = 0 ( * )

Ta coù (*) : ⇔ ( 4 cos3 x − 3 cos x ) − 4 ( 2 cos2 x − 1) + 3 cos x − 4 = 0

⇔ 4 cos3 x − 8 cos2 x = 0 ⇔ 4 cos2 x ( cos x − 2 ) = 0 ⇔ cos x = 0 hay cos x = 2 ( loaïi vì cos x ≤ 1) ⇔ x=

π + kπ ( k ∈ Z ) 2

π + kπ ≤ 14 2 π π 1 14 1 − ≈ 3, 9 ⇔ − ≤ kπ ≤ 14 − ⇔ −0, 5 = − ≤ k ≤ 2 2 2 π 2 ⎧ π 3π 5π 7π ⎫ Maø k ∈ Z neâ n k ∈ {0,1, 2, 3} . Do ñoù : x ∈ ⎨ , , , ⎬ ⎩2 2 2 2 ⎭ Ta coù : x ∈ [ 0,14] ⇔ 0 ≤

Baø i 29 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i D, naê m 2004) Giaû i phöông trình : ( 2 cos x − 1)( 2 sin x + cos x ) = sin 2x − sin x ( *)

Ta coù (*) ⇔ ( 2 cos x − 1)( 2 sin x + cos x ) = sin x ( 2 cos x − 1)

⇔ ( 2 cos x − 1) ⎡⎣( 2 sin x + cos x ) − sin x ⎤⎦ = 0 ⇔ ( 2 cos x − 1)( sin x + cos x ) = 0

1 ∨ sin x = − cos x 2 π ⎛ π⎞ ⇔ cos x = cos ∨ tgx = −1 = tg ⎜ − ⎟ 3 ⎝ 4⎠ π π ⇔ x = ± + k2π ∨ x = − + kπ, ( k ∈ Z ) 3 4 ⇔ cos x =

Baø i 30 : Giaû i phöông trình cos x + cos 2x + cos 3x + cos 4x = 0 (*) Ta coù (*) ⇔ ( cos x + cos 4x ) + ( cos 2x + cos 3x ) = 0

5x 3x 5x x .cos + 2 cos .cos = 0 2 2 2 2 5x ⎛ 3x x⎞ 2 cos + cos ⎟ = 0 ⎜ cos 2 ⎝ 2 2⎠ 5x x 4 cos cos x cos = 0 2 2 5x x = 0 ∨ cos x = 0 ∨ cos = 0 cos 2 2 5x π π x π = + kπ ∨ x = + kπ ∨ = + kπ 2 2 2 2 2 π 2kπ π x= + ∨ x = + kπ ∨ x = π + 2π, ( k ∈ Z ) 5 5 2

⇔ 2 cos ⇔ ⇔ ⇔ ⇔ ⇔

Baø i 31: Giaûi phöông trình sin 2 x + sin 2 3x = cos2 2x + cos2 4x ( * )

1 1 1 1 (1 − cos 2x ) + (1 − cos 6x ) = (1 + cos 4x ) + (1 + cos 8x ) 2 2 2 2 ⇔ − ( cos 2x + cos 6x ) = cos 4x + cos 8x Ta coù (*) ⇔

⇔ −2 cos 4x cos 2x = 2 cos 6x cos 2x ⇔ 2 cos 2x ( cos 6x + cos 4x ) = 0 ⇔ 4 cos 2x cos 5x cos x = 0 ⇔ cos 2x = 0 ∨ cos 5x = 0 ∨ cos x = 0 π π π ⇔ 2x = + kπ ∨ 5x + kπ ∨ x = + kπ, k ∈ 2 2 2 π kπ π kπ π ∨x= + ∨ x = + kπ , k ∈ ⇔ x= + 4 2 10 5 2 Baø i 32 : Cho phöông trình ⎛π x⎞ 7 sin x.cos 4x − sin 2 2x = 4 sin 2 ⎜ − ⎟ − ( *) ⎝4 2⎠ 2 Tìm caù c nghieä m cuû a phöông trình thoû a : x − 1 < 3

1 ⎡ π ⎤ 7 (1 − cos 4x ) = 2 ⎢1 − cos ⎛⎜ − x ⎞⎟ ⎥ − 2 ⎝2 ⎠⎦ 2 ⎣ 1 1 3 sin x cos 4x − + cos 4x = − − 2sin x 2 2 2 1 sin x cos 4x + cos 4x + 1 + 2sin x = 0 2 1⎞ 1⎞ ⎛ ⎛ cos 4x ⎜ sin x + ⎟ + 2 ⎜ sin x + ⎟ = 0 2⎠ 2⎠ ⎝ ⎝ 1 ( cos 4x + 2) ⎜⎛ sin x + ⎟⎞ = 0 2⎠ ⎝ π ⎡ ⎡cos 4x = −2 ( loaïi ) x = − + k 2π ⎢ 6 ⎢ ⎢sin x = − 1 = sin ⎛ − π ⎞ ⇔ ⎢ ⎢ x = 7π + 2hπ ⎜ ⎟ ⎢⎣ 2 ⎝ 6⎠ ⎢⎣ 6 coù : x − 1 < 3 ⇔ −3 < x − 1 < 3 ⇔ −2 < x < 4

Ta coù : (*)⇔ sin x.cos 4x −

⇔ ⇔ ⇔ ⇔

⇔ Ta

π + k2π < 4 6 π π 1 1 2 1 − <k< + ⇔ − 2 < 2kπ < 4 + ⇔ 6 6 12 π π 12 π Do k ∈ Z neâ n k = 0. Vaä y x = − 6 7π −2 < + h2π < 4 6 7π 7π 1 7 2 7 < h2π < 4 − ⇔− − <h< − ⇔ −2 − 6 6 π 12 π 12 7π −π 7π hay x = .Toù m laï i x = ⇒h = 0 ⇒ x = 6 6 6 1 −π + kπ, k ∈ Caù c h khaù c : sin x = − ⇔ x = (−1)k 2 6 −π −2 −1 4 + kπ < 4 ⇔ < (−1)k +k< Vaä y : −2 < (−1)k 6 6 π π −π 7π hay x = ⇔ k=0 vaø k = 1. Töông öù n g vôù i x = 6 6 Vaä y : −2 < −

Baø i 33 : Giaû i phöông trình sin 3 x cos 3x + cos3 x sin 3x = sin 3 4x ( * ) Ta coù : (*)⇔ sin 3 x ( 4 cos3 x − 3 cos x ) + cos3 x ( 3sin x − 4 sin 3 x ) = sin 3 4x

⇔ 4 sin3 x cos3 x − 3sin3 x cos x + 3sin x cos3 x − 4 sin3 x cos3 x = sin3 4x ⇔ 3sin x cos x ( cos2 x − sin 2 x ) = sin 3 4x ⇔

3 sin 2x cos 2x = sin3 4x 2

3 sin 4x = sin3 4x 4 ⇔ 3sin 4x − 4 sin3 4x = 0 ⇔ sin12x = 0 ⇔

kπ ( k ∈ Z) 12 Baø i 34 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i B, naê m 2002) Giaû i phöông trình : sin 2 3x − cos2 4x = sin 2 5x − cos2 6a ( * ) ⇔ 12x = kπ

⇔ x=

Ta coù : (*)⇔ 1 1 1 1 (1 − cos 6x ) − (1 + cos 8x ) = (1 − cos10x ) − (1 + cos12x ) 2 2 2 2 ⇔ cos 6x + cos 8x = cos10x + cos12x ⇔ 2 cos7x cos x = 2 cos11x cos x ⇔ 2 cos x ( cos 7x − cos11x ) = 0

⇔ cos x = 0 ∨ cos7x = cos11x π ⇔ x = + kπ ∨ 7x = ±11x + k 2π 2 π kπ kπ ∨x= ,k ∈ ⇔ x = + kπ ∨ x = − 2 2 9 Baø i 35 : Giaû i phöông trình ( sin x + sin 3x ) + sin 2x = ( cos x + cos 3x ) + cos 2x ⇔ 2sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x ⇔ sin 2x ( 2 cos x + 1) = cos 2x ( 2 cos x + 1) ⇔ ( 2 cos x + 1) ( sin 2x − cos 2x ) = 0

1 2π = cos ∨ sin 2x = cos 2x 2 3 2π π + k2π ∨ tg2x = 1 = tg ⇔ x=± 3 4 2π π π + k2π ∨ x = + k , ( k ∈ Z ) ⇔ x=± 3 8 2 ⇔ cos x = −

Baø i 36: Giaû i phöông trình cos 10x + 2 cos2 4x + 6 cos 3x. cos x = cos x + 8 cos x. cos3 3x ( * ) Ta coù : (*)⇔ cos10x + (1 + cos 8x ) = cos x + 2 cos x ( 4 cos3 3x − 3 cos 3x )

⇔ ( cos10x + cos 8x ) + 1 = cos x + 2 cos x.cos 9x ⇔ 2 cos 9x cos x + 1 = cos x + 2 cos x.cos 9x ⇔ cos x = 1 ⇔ x = k2π ( k ∈ Z ) Baø i 37 : Giaû i phöông trình

4 sin 3 x + 3 cos3 x − 3sin x − sin 2 x cos x = 0 ( * )

Ta coù : (*) ⇔ sin x ( 4 sin 2 x − 3) − cos x ( sin 2 x − 3 cos2 x ) = 0

⇔ sin x ( 4 sin 2 x − 3) − cos x ⎡⎣sin 2 x − 3 (1 − sin 2 x ) ⎤⎦ = 0 ⇔ ( 4 sin 2 x − 3) ( sin x − cos x ) = 0 ⇔ ⎡⎣ 2 (1 − cos 2x ) − 3⎤⎦ ( sin x − cos x ) = 0 1 2π ⎡ cos 2x cos = − = ⇔ ⎢ 2 3 ⎢ ⎣sin x = cos x

2π ⎡ 2x = ± + k2π ⎢ ⇔ 3 ⎢ ⎣ tgx = 1

π ⎡ x = ± + kπ ⎢ 3 ⇔ ⎢ ⎢ x = π + kπ ⎢⎣ 4

( k ∈ Z)

Baø i 38 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i B naê m 2005) Giaû i phöông trình : sin x + cos x + 1 + sin 2x + cos 2x = 0 ( * ) Ta coù : (*) ⇔ sin x + cos x + 2sin x cos x + 2 cos2 x = 0 ⇔ sin x + cos x + 2 cos x ( sin x + cos x ) = 0

⇔ ( sin x + cos x ) (1 + 2 cos x ) = 0

⎡sin x = − cos x ⇔ ⎢ ⎢cos 2x = − 1 = cos 2π 2 3 ⎣ ⎡ tgx = −1 ⇔ ⎢ ⎢ x = ± 2π + k 2π 3 ⎣ π ⎡ ⎢ x = − 4 + kπ ⇔ ⎢ ( k ∈ Z) ⎢ x = ± 2π + k2π ⎢⎣ 3 Baø i 39 : Giaû i phöông trình ( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + 4 cos2 x = 3 ( *) Ta coù : (*) ⇔ ( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + 4 (1 − sin 2 x ) − 3 = 0

⇔ ( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + (1 + 2 sin x )(1 − 2 sin x ) = 0 ⇔ ( 2 sin x + 1) ⎡⎣ 3 cos 4x + 2 sin x − 4 + (1 − 2 sin x ) ⎤⎦ = 0 ⇔ 3 ( cos 4x − 1)( 2 sin x + 1) = 0 ⇔ cos 4x = 1 ∨ sin x = −

1 ⎛ π⎞ = sin ⎜ − ⎟ 2 ⎝ 6⎠

π 7π + k2π ∨ x = + k2π 6 6 kπ π 7π ∨ x = − + k2π ∨ x = + k2π, ( k ∈ Z) ⇔ x= 2 6 6 ⇔ 4x = k2π ∨ x = −

Baø i 40: Giaû i phöông trình sin 6 x + cos6 x = 2 ( sin 8 x + cos8 x ) ( * ) Ta coù : (*) ⇔ sin6 x − 2sin8 x + cos6 x − 2 cos8 x = 0 ⇔ sin 6 x (1 − 2 sin 2 x ) − cos6 x ( 2 cos2 x − 1) = 0

⇔ sin6 x cos 2x − cos6 x. cos 2x = 0 ⇔ cos 2x ( sin 6 x − cos6 x ) = 0 ⇔ cos 2x = 0 ∨ sin6 x = cos6 x ⇔ cos 2x = 0 ∨ tg 6 x = 1 π ⇔ 2x = ( 2k + 1) ∨ tgx = ±1 2 π π ⇔ x = ( 2k + 1) ∨ x = ± + kπ 4 4 π kπ ⇔ x= + ,k ∈ 4 2 Baø i 41 : Giaû i phöông trình

1 ( *) 16 Ta thaá y x = kπ khoâ n g laø nghieä m cuû a (*) vì luù c ñoù cos x = ±1, cos 2x = cos 4x = cos 8x = 1 1 (*) thaøn h : ±1 = voâ nghieä m 16 Nhaâ n 2 veá cuû a (*) cho 16sin x ≠ 0 ta ñöôï c (*) ⇔ (16 sin x cos x ) cos 2x.cos 4x.cos 8x = sin x vaø sin x ≠ 0

cos x.cos 2x.cos 4x.cos 8x =

⇔ ( 8 sin 2x cos 2x ) cos 4x.cos 8x = sin x vaø sin x ≠ 0 ⇔ ( 4 sin 4x cos 4x ) cos 8x = sin x vaø sin x ≠ 0 ⇔ 2sin 8x cos 8x = sin x vaø sin x ≠ 0 ⇔ sin16x = sin x vaø sin x ≠ 0 k2π π kπ ∨x= + , ( k ∈ Z) ⇔x = 15 17 17 Do : x = hπ khoâ n g laø nghieä m neâ n k ≠ 15m vaø 2k + 1 ≠ 17n ( n, m ∈ Z ) 3 Baø i 42: Giaû i phöông trình 8cos ⎛⎜ x +

Ñaët t = x +

π π ⇔x=t− 3 3

π⎞

3⎠

= cos 3x ( * )

Thì cos 3x = cos ( 3t − π ) = cos ( π − 3t ) = − cos 3t Vaä y (*) thaø n h 8 cos3 t = − cos 3t ⇔ 8 cos3 t = −4 cos3 t + 3 cos t ⇔ 12 cos3 t − 3 cos t = 0 ⇔ 3 cos t ( 4 cos2 t − 1) = 0

⇔ 3 cos t ⎡⎣2 (1 + cos 2t ) − 1⎤⎦ = 0 ⇔ cos t ( 2 cos 2t + 1) = 0

1 2π = cos 2 3 π 2π + k2π ⇔ t = ( 2k + 1) ∨ 2t = ± 2 3 π π ⇔ t = + kπ ∨ t = ± + kπ 2 3 π Maø x = t − 3 π 2π + kπ, ( vôùik ∈ Z ) Vaä y (*) ⇔ x = + k2π ∨ x = kπ ∨ x = 6 3 Ghi chuù : Khi giaû i caù c phöông trình löôï n g giaù c coù chöù a tgu, cotgu, coù aå n ôû maã u , hay chöù a caê n baä c chaü n ... ta phaû i ñaë t ñieà u kieä n ñeå phöông trình xaù c ñònh. Ta seõ duø n g caù c caù c h sau ñaâ y ñeå kieå m tra ñieà u kieä n xem coù nhaä n nghieä m hay khoâ n g. + Thay caùc giaù trò x tìm ñöôï c vaø o ñieà u kieä n thöû laï i xem coù thoû a Hoaë c + Bieå u dieã n caù c ngoï n cung ñieà u kieä n vaø caù c ngoï n cung tìm ñöôïc treâ n cuø n g moä t ñöôø n g troø n löôï n g giaù c . Ta seõ loaï i boû ngoï n cung cuû a nghieä m khi coù truø n g vôù i ngoï n cung cuû a ñieà u kieä n . Hoaë c + So vôi caù c ñieà u kieä n trong quaù trình giaûi phöông trình. ⇔ cos t = 0 ∨ cos 2t = −

Baø i 43 : Giaû i phöông trình tg 2 x − tgx.tg3x = 2 ( * ) π hπ ⎧cos x ≠ 0 ⇔ cos3x ≠ 0 ⇔ x ≠ + Ñieà u kieä n ⎨ 3 6 3 ⎩cos 3x = 4 cos x − 3 cos x ≠ 0 Luù c ñoù ta coù (*) ⇔ tgx ( tgx − tg3x ) = 2

sin x ⎛ sin x sin 3x ⎞ − ⎜ ⎟=2 cos x ⎝ cos x cos 3x ⎠ ⇔ sin x ( sin x cos 3x − cos x sin 3x ) = 2 cos2 x cos 3x

⇔ sin x sin ( −2x ) = 2 cos2 x. cos 3x ⇔ −2 sin2 x cos x = 2 cos2 x cos 3x ⇔ − sin2 x = cos x cos 3x (do cos x ≠ 0 ) 1 1 ⇔ − (1 − cos 2x ) = ( cos 4x + cos 2x ) 2 2 ⇔ cos 4x = −1 ⇔ 4x = π + k2π

π kπ + ( k ∈ Z) 4 2 so vôù i ñieà u kieä n π kπ 2 ⎛ 3π 3kπ ⎞ + ≠ 0 ( nhaän ) Caù c h 1 : Khi x = + thì cos 3x = cos ⎜ ⎟=± 2 ⎠ 2 4 2 ⎝ 4 Caù c h 2 : Bieå u dieã n caù c ngoï n cung ñieà u kieä n vaø ngoï n cung nghieä m ta thaá y khoâ n g coù ngoï n cung naø o truø n g nhau. Do ñoù : π kπ (*) ⇔ x = + 4 2 Löu yù caù c h 2 raá t maá t thôøi gian Caù c h 3 : 3π 3kπ π + = + hπ Neá u 3x = 4 2 2 Thì 3 + 6k = 2 + 4h ⇔ 1 = 4h − 6k 1 ⇔ = 2h − 3k (voâ lyù vì k, h ∈ Z ) 2 ⇔x =

Baø i 44: Giaûi phöông trình

11 ( *) 3 ⎧cos x ≠ 0 ⎪ Ñieà u kieä n ⎨sin x ≠ 0 ⇔ sin 2x ≠ 0 ⎪sin 2x ≠ 0 ⎩

tg 2 x + cot g 2 x + cot g 2 2x =

Do ñoù : ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 11 − 1⎟ + ⎜ − 1⎟ + ⎜ − 1⎟ = (*) ⇔ ⎜ 2 2 2 3 ⎝ cos x ⎠ ⎝ sin x ⎠ ⎝ sin 2x ⎠ 1 1 1 20 + + = ⇔ 2 2 2 2 cos x sin x 4 sin x cos x 3 2 2 4 sin x + 4 cos x + 1 20 = ⇔ 4 sin2 x cos2 x 3 5 20 = ⇔ sin2 2x 3 3 ⇔ sin2 2x = (nhaä n do sin2x ≠ 0 ) 4 1 3 ⇔ (1 − cos 4x ) = 2 4 1 2π ⇔ cos 4x = − = cos 2 3 2π + k2π ⇔ 4x = ± 3 π kπ ⇔x = ± + ( k ∈ Z) 6 2

Chuù yù : Coù theå deã daø n g chöù n g minh : tgx + cot gx = 2 ⎛ 1 ⎞ 11 Vaä y (*) ⇔ ( tgx + cot gx ) − 2 + ⎜ − 1⎟ = 2 3 ⎝ sin x ⎠ 5 20 = ⇔ 2 sin 2x 3

2 sin 2x

Baø i 45 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i D, naê m 2003) Giaû i phöông trình x ⎛x π⎞ sin 2 ⎜ − ⎟ tg 2 x − cos2 = 0 ( *) 2 ⎝2 4⎠ Ñieà u kieä n : cos x ≠ 0 ⇔ sin x ≠ ±1 luù c ñoù : 1⎡ π ⎞ ⎤ sin 2 x 1 ⎛ − [1 + cos x ] = 0 (*) ⇔ ⎢1 − cos ⎜ x − ⎟ ⎥ 2⎣ 2 ⎠ ⎦ cos2 x 2 ⎝

(1 − sin x ) (1 − cos2 x )

− (1 + cos x ) = 0 1 − sin 2 x 1 − cos2 x − (1 + cos x ) = 0 ⇔ 1 + sin x ⎡ 1 − cos x ⎤ − 1⎥ = 0 ⇔ (1 + cos x ) ⎢ ⎣ 1 + sin x ⎦ ⇔ (1 + cos x ) ( − cos x − sin x ) = 0 ⇔

⎡cos x = −1 ( nhaändo cos x ≠ 0 ) ⇔ ⎢ ⎣ tgx = −1 ⎡ x = π + k2π ⇔ ⎢ ⎢ x = − π + kπ ⎣ 4 Baø i 46 : Giaû i phöông trình sin 2x ( cot gx + tg2x ) = 4 cos2 x ( * )

⎧sin x ≠ 0 Ñieà u kieä n : ⎨ ⎩cos 2x ≠ 0

⎧sin x ≠ 0 ⎨ 2 ⎩2 cos x − 1 ≠ 0

cos x sin 2x + sin x cos 2x cos 2x cos x + sin 2x sin x = sin x cos 2x cos x = sin x cos 2x cos x ⎛ ⎞ 2 Luù c ñoù : (*) ⇔ 2 sin x cos x ⎜ ⎟ = 4 cos x ⎝ sin x cos 2x ⎠ Ta coù : cot gx + tg2x =

⎧cos x ≠ ±1 ⎪ ⎨ 2 ⎪cos x ≠ ± 2 ⎩

2 cos2 x = 4 cos2 x ( Do sin x ≠ 0 ) ⇔ cos 2x ⎡ ⎛ ⎞ 2 vaø ≠ ±1 ⎟⎟ ⎡cos x = 0 ⎢cos x = 0 ⎜⎜ Nhaän do cos x ≠ 2 ⎝ ⎠ ⇔ ⎢ 1 ⇔ ⎢ ⎢ ⎢ =2 1 π ⎣ cos 2x ⎢cos 2x = = cos , ( nhaän do sin x ≠ 0) 2 3 ⎣ π ⎡ ⎢ x = 2 + kπ ⇔ ⎢ ( k ∈ Z) ⎢ x = ± π + kπ ⎢⎣ 6 Baø i 47 : Giaû i phöông trình: cot g 2 x − tg 2 x = 16 (1 + cos 4x ) cos 2x cos2 x sin 2 x − Ta coù : cot g 2 x − tg 2 x = sin2 x cos2 x cos4 x − sin4 x 4 cos 2x = = sin2 x cos2 x sin2 2x ⎧sin 2x ≠ 0 Ñieà u kieä n : ⎨ ⇔ sin 4x ≠ 0 ⎩cos 2x ≠ 0 4 = 16 (1 + cos 4x ) Luù c ñoù (*) ⇔ sin2 2x ⇔ 1 = 4 (1 + cos 4x ) sin2 2x

⇔ 1 = 2 (1 + cos 4x ) (1 − cos 4x )

(

)

⇔ 1 = 2 1 − cos2 4x = 2 sin 2 4x 1 ( nhaän do sin 4x ≠ 0) 2 1 1 ⇔ (1 − cos 8x ) = 2 2 π kπ ⇔ cos 8x = 0 ⇔ x = + ,k ∈ 16 8 ⇔ sin2 4x =

Baø i 48: Giaûi phöông trình: sin 4 x + cos4 x =

⎧ π⎞ ⎛ ⎪sin ⎜ x + 3 ⎟ ≠ 0 ⎪ ⎝ ⎠ Ñieà u kieä n ⎨ ⇔ ⎪sin ⎛ π − x ⎞ ≠ 0 ⎜ ⎟ ⎪⎩ ⎝6 ⎠

⎧ ⎛ ⎪sin ⎜ x + ⎪ ⎝ ⎨ ⎪cos ⎛ x + ⎜ ⎪⎩ ⎝

π⎞ 7 ⎛ ⎛π ⎞ cot g ⎜ x + ⎟ cot g ⎜ − x ⎟ ( *) 8 3⎠ ⎝ ⎝6 ⎠

π⎞ ⎟≠0 3⎠ π⎞ ⎟≠0 3⎠

2π ⎞ ⎛ ⇔ sin ⎜ 2x + ⎟≠0 3 ⎠ ⎝

1 3 ⇔ − sin 2x + cos 2x ≠ 0 2 2 ⇔ tg2x ≠ 3

(

Ta coù : sin4 x + cos4 x = sin2 x + cos2 x

)

2

− 2sin2 x.cos2 x = 1 −

1 sin2 2x 2

π⎞ π⎞ ⎛π ⎛ ⎛π ⎞ ⎛ ⎞ Vaø : cot g ⎜ x + ⎟ .cot g ⎜ − x ⎟ = cot g ⎜ x + ⎟ .tg ⎜ + x ⎟ = 1 3⎠ 3⎠ ⎝3 ⎝ ⎝6 ⎠ ⎝ ⎠ 1 7 Luù c ñoù : (*) ⇔ 1 − sin2 2x = 2 8 1 1 ⇔ − (1 − cos 4x ) = − 4 8 1 ⇔ cos 4x = 2 π π kπ ⇔ 4x = ± + k2π ⇔ x = ± + 3 12 2 3 (nhaä n do tg2x = ± ≠ 3) 3

Baø i 49: Giaû i phöông trình 2tgx + cot g2x = 2 sin 2x +

1 ( *) sin 2x

⎧cos 2x ≠ 0 Ñieà u kieä n : ⎨ ⇔ sin 2x ≠ 0 ⇔ cos 2x ≠ ±1 ⎩sin 2x ≠ 0 2 sin x cos 2x 1 + = 2 sin 2x + Luù c ñoù : (*) ⇔ cos x sin 2x sin 2x 2 2 ⇔ 4 sin x + cos 2x = 2 sin 2x + 1

(

)

⇔ 4 sin2 x + 1 − 2 sin 2 x = 8 sin2 x cos2 x + 1

(

)

⇔ 2 sin2 x 1 − 4 cos2 x = 0 ⇔ 2 sin2 x ⎡⎣1 − 2 (1 + cos 2x ) ⎤⎦ = 0 ⎡sin x = 0 ( loaïi do sin 2x ≠ 0 ⇒ sin x ≠ 0 ) ⇔⎢ ⎢cos 2x = − 1 = cos 2π ( nhaän do cos 2x ≠ ±1) ⎢⎣ 2 3 2π ⇔ 2x = ± + k2π ( k ∈ Z ) 3 π ⇔ x = ± + kπ, k ∈ 3

Baø i 51: Giaû i phöông trình:

3 ( sin x + tgx ) tgx − sin x

− 2 (1 + cos x ) = 0 ( *)

sin x − sin x ≠ 0 cos x ⎧sin x ≠ 0 sin x (1 − cos x ) ⎪ ≠ 0 ⇔ ⎨cos x ≠ 0 ⇔ sin 2x ≠ 0 ⇔ cos x ⎪cos x ≠ 1 ⎩ Ñieà u kieä n : tgx − sin x ≠ 0 ⇔

Luù c ñoù (*)⇔

3 ( sin x + tgx ) .cot gx − 2 (1 + cos x ) = 0 ( tgx − sin x ) .cot gx

3 ( cos x + 1) − 2 (1 + cos x ) = 0 (1 − cos x )

3 − 2 = 0 ( do sin x ≠ 0 neân cos x + 1 ≠ 0) 1 − cos x ⇔ 1 + 2 cos x = 0 1 ⇔ cos x = − (nhaä n so vôù i ñieà u kieä n ) 2 2π + k2π, k ∈ ⇔ x=± 3 Baø i 52 : Giaû i phöông trình 2 2 (1 − cos x ) + (1 + cos x ) − tg 2 x sin x = 1 1 + sin x + tg 2 x * ( ) ( ) 4 (1 − sin x ) 2 ⇔

⎧cos x ≠ 0 Ñieà u kieä n : ⎨ ⇔ cos x ≠ 0 ⎩sin x ≠ 1 2 (1 + cos2 x ) sin 3 x 1 sin 2 x − = 1 + sin x + Luù c ñoù (*)⇔ ( ) 4 (1 − sin x ) 1 − sin 2 x 2 1 − sin 2 x

⇔ (1 + cos2 x ) (1 + sin x ) − 2 sin 3 x = (1 + sin x ) (1 − sin 2 x ) + 2 sin 2 x ⇔ (1 + sinx ) (1 + cos2 x ) = (1 + sin x ) cos2 x + 2 sin 2 x (1 + sin x ) ⎡1 + sin x = 0 ⇔ ⎢ 2 2 2 ⎣1 + cos x = cos x + 2 sin x ⎡sin x = −1 ( loaïi do cos x ≠ 0 ) ⇔ ⎢ ⇔ cos2x = 0 ⎣1 = 1 − cos 2x π ⇔ 2x = + kπ 2 π π ⇔ x = + k (nhaä n do cosx ≠ 0) 4 2

Baø i 53 : Giaû i phöông trình Ñieà u kieä n cos 5x ≠ 0 Luù c ñoù : (*) ⇔ cos 3x.

cos 3x.tg5x = sin 7x ( * )

sin 5x = sin 7x cos 5x

⇔ sin 5x.cos 3x = sin 7x.cos 5x 1 1 ⇔ [sin 8x + sin 2x ] = [sin12x + sin 2x ] 2 2 ⇔ sin 8x = sin12x ⇔ 12x = 8x + k2π ∨ 12x = π − 8x + k2π kπ π kπ ∨ x= + ⇔x = 2 20 10 So laï i vôù i ñieà u kieä n kπ 5kπ kπ x= thì cos 5x = cos = cos (loaï i neá u k leû ) 2 2 2 kπ π ⎛ π kπ ⎞ x= thì cos 5x = cos ⎜ + + ⎟ ≠ 0 nhaän 2 ⎠ 20 10 ⎝4 π kπ + Do ñoù : (*)⇔ x = hπ ∨ x = , vôù i k, h ∈ 20 10 Baø i 54 : Giaû i phöông trình sin4 x + cos4 x 1 = ( tgx + cot g2x ) ( *) sin 2x 2 Ñieà u kieä n : sin 2x ≠ 0

Ta coù : sin 4 x + cos4 x = ( sin 2 x + cos2 x ) − 2 sin 2 x cos2 x 2

=1−

1 sin2 2x 2

sin x cos 2x + cos x sin 2x sin 2x sin x + cos x cos 2x = cos x sin 2x cos ( 2x − x ) 1 = = cos x sin 2x sin 2x 1 1 − sin 2 2x 1 2 Do ñoù : (*) ⇔ = sin 2x 2 sin 2x 1 1 ⇔ 1 − sin 2 2x = 2 2 2 ⇔ sin 2x = 1 ( nhaän do sin 2x ≠ 0 ) tgx + cot g2x =

⇔ cos2 2x = 0 π + kπ, k ∈ 2 π kπ , k ∈ ⇔x = + 4 2 ⇔ 2x =

Baø i 55 : Giaû i phöông trình tg 2 x.cot g 2 2x.cot g3x = tg 2 x − cot g 2 2x + cot g3x ( * ) Ñieà u kieä n : cos x ≠ 0 ∧ sin 2x ≠ 0 ∧ sin 3x ≠ 0

⇔ sin 2x ≠ 0 ∧ sin 3x ≠ 0

Luùc ñoù (*) ⇔ cotg3x ( tg 2 x cot g 2 2x − 1) = tg 2 x − cot g 2 2x ⎡⎛ 1 − cos 2x ⎞ ⎛ 1 + cos 4x ⎞ ⎤ 1 − cos 2x 1 + cos 4x ⇔ cot g3x ⎢⎜ − ⎟⎜ ⎟ − 1⎥ = ⎣⎝ 1 + cos 2x ⎠ ⎝ 1 − cos 4x ⎠ ⎦ 1 + cos 2x 1 − cos 4x

⇔ cot g3x ⎡⎣(1 − cos 2x )(1 + cos 4x ) − (1 + cos 2x )(1 − cos 4x ) ⎤⎦ = (1 − cos 2x )(1 − cos 4x ) − (1 + cos 4x )(1 + cos 2x )

⇔ cot g3x [ 2 cos 4x − 2 cos 2x ] = −2 ( cos 4x + cos 2x ) cos 3x [ −4 sin 3x sin x] = −4 cos 3x cos x sin 3x ⇔ cos 3x sin x = cos 3x cos x ( do sin 3x ≠ 0) ⇔

⇔ cos 3x = 0 ∨ sin x = cos x π + kπ ∨ tgx = 1 2 π kπ π ⇔x= + ∨ x = + lπ ( k, l ∈ Z ) 6 3 4 So vôù i ñieà u kieä n : sin 2x.sin 3x ≠ 0 π kπ ⎛ π 2kπ ⎞ ⎛π ⎞ * Khi x = + thì sin ⎜ + ⎟ .sin ⎜ + kπ ⎟ ≠ 0 3 ⎠ 6 3 ⎝3 ⎝2 ⎠ ⎛ 1 + 2k ⎞ ⇔ sin ⎜ ⎟π ≠ 0 ⎝ 3 ⎠ Luoâ n ñuù n g ∀ k thoûa 2k + 1 ≠ 3m ( m ∈ Z ) ⇔ 3x =

* Khi x =

π 2 ⎛π ⎞ ⎛ 3π ⎞ + lπ thì sin ⎜ + 2lπ ⎟ sin ⎜ + 3lπ ⎟ = ± ≠0 2 4 ⎝2 ⎠ ⎝ 4 ⎠

luoâ n ñuù n g

π kπ ⎡ ⎢ x = 6 + 3 , k ∈ Z ∧ 2k ≠ 3m − 1 ( m ∈ ) Do ñoù : (*) ⇔ ⎢ ⎢ x = π + lπ, l ∈ ⎢⎣ 4 Caù c h khaù c: (*) ⇔ cotg3x ( tg 2 x cot g 2 2x − 1) = tg 2 x − cot g 2 2x

tg 2 x − cot g 2 2x tg 2 2x.tg 2 x − 1 = tg 2 x cot g 2 2x − 1 tg 2 x − tg 2 2x (1 + tg2x.tgx ) (1 − tg2x.tgx ) ⇔ cot g3x = (tg2x − tgx) ( tg2x + tgx) ⇔ cot g3x = cot gx. cotg3x ⇔ cos 3x = 0 ∨ sin x = cos x

⇔ cot g3x =

BAØI TAÄP

1.

2.

3.

⎛π ⎞ Tìm caù c nghieä m treâ n ⎜ , 3π ⎟ cuû a phöông trình: ⎝3 ⎠ 5π ⎞ 7π ⎞ ⎛ ⎛ sin ⎜ 2x + ⎟ − 3 cos ⎜ x − ⎟ = 1 + 2 sin x 2 ⎠ 2 ⎠ ⎝ ⎝ ⎛ π⎞ Tìm caù c nghieä m x treâ n ⎜ 0, ⎟ cuû a phöông trình ⎝ 2⎠ 2 2 sin 4x − cos 6x = sin (10, 5π + 10x ) Giaû i caù c phöông trình sau: a/ sin 3 x + cos3 x = 2 sin5 x + cos5 x

(

)

sin x + sin 2x + sin 3x = 3 cos x + cos 2x + cos 3x 1 + cos x c/ tg 2 x = 1 − sin x d/ tg2x − tg3x − tg5x = tg2x.tg3x.tg5x 4 e/ cos x = cos2 x 3 π⎞ 1 1 ⎛ + f/ 2 2 sin ⎜ x + ⎟ = 4 ⎠ sin x cos x ⎝ 2 i/ 2tgx + cot g2x = 3 + sin 2x 2 h/ 3tg3x + cot g2x = 2tgx + sin 4x 2 2 2 k/ sin x + sin 2x + sin 3x = 2 sin 2x + 2 cos x = 0 l/ 1 + sin x b/

m/

25 − 4x 2 ( 3sin 2πx + 8 sin πx ) = 0

sin x.cot g5x =1 cos 9x 2 = 2tg2x − cot g4x o/ 3tg6x − sin 8x p/ 2 sin 3x 1 − 4 sin 2 x = 1 n/

(

q/ tg 2 x =

)

1 + cos x 1 − sin x

r/ cos3 x cos 3x + sin 3 x sin 3x =

2 4

⎛x⎞ ⎛x⎞ 5 s/ sin4 ⎜ ⎟ + cos4 ⎜ ⎟ = ⎝ 3⎠ ⎝ 3⎠ 8 3 3 t/ cos x − 4 sin x − 3 cos x sin2 x + sin x = 0 x x u/ sin4 + cos4 = 1 − 2sin x 2 2

π⎞ π⎞ ⎛ ⎛ v/ sin ⎜ 3x − ⎟ = sin 2x.sin ⎜ x + ⎟ 4⎠ 4⎠ ⎝ ⎝

( 2 − sin x ) sin 3x w/ tg x + 1 = 2

4

4.

cos4 x

x ⎛ ⎞ y/ tgx + cos x − cos2 x = sin x ⎜ 1 + tg tgx ⎟ 2 ⎝ ⎠ Cho phöông trình: ( 2 sin x − 1)( 2 cos 2x + 2 sin x + m ) = 3 − 4 cos2 x (1)

a/ Giaû i phöông trình khi m = 1 b/ Tìm m ñeå (1) coù ñuù n g 2 nghieä m treâ n [ 0, π ] 5.

( ÑS: m = 0 ∨ m < −1 ∨ m > 3 ) Cho phöông trình: 4 cos5 x sin x − 4 sin5 x.cos x = sin2 4x + m (1) Bieá t raè n g x = π laø moä t nghieä m cuû a (1). Haõ y giaû i phöông trình trong tröôø n g hôï p ñoù .

Th.S Phạm Hồng Danh TT luyện thi Đại học CLC Vĩnh Viễn

Pt lg coban