Phuong trinh bac hai doi voi mot hslg

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					LƯỢNG GIÁC
CHÖÔNG III: PHÖÔNG TRÌNH BAÄ C HAI VÔÙ I CAÙ C HAØ M SOÁ LÖÔÏ N G GIAÙ C

a sin2 u + b sin u + c = 0 a cos2 u + b cos u + c = 0 atg 2 u + btgu = c = 0 a cot g 2 u + b cot gu + c = 0

( a ≠ 0) ( a ≠ 0) ( a ≠ 0) ( a ≠ 0)

Caù c h giaû i: t = sin u hay t = cos u vôù i t ≤ 1 Ñaët :

π + kπ ) 2 t = cot gu (ñieà u kieä n u ≠ kπ )
t = tgu (ñieà u kieä n u ≠

Caù c phöông trình treâ n thaø n h: at 2 + bt + c = 0 Giaû i phöông trình tìm ñöôïc t, so vôù i ñieà u kieä n ñeå nhaä n nghieä m t. Töø ñoù giaû i phöông trình löôï n g giaù c cô baû n tìm ñöôï c u.

Baø i 56: (Ñeà thi tuyeån sinh Ñaï i hoï c khoá i A, naê m 2002) Tìm caù c nghieä m treâ n ( 0, 2π ) cuû a phöông trình
cos 3x + sin 3x ⎞ ⎛ 5 ⎜ sin x + ⎟ = 3 + cos 2x ( * ) 1 + 2 sin 2x ⎠ ⎝ 1 Ñieà u kieä n : sin 2x ≠ − 2 Ta coù : sin 3x + cos 3x = 3sin x − 4 sin 3 x + 4 cos3 x − 3 cos x
= −3 ( cos x − sin x ) + 4 cos3 x − sin3 x

(

(

)

) (

)

= ( cos x − sin x ) ⎡ −3 + 4 cos2 x + cos x sin x + sin 2 x ⎤ ⎣ ⎦ = ( cos x − sin x )(1 + 2 sin 2x )

(

)

Luù c ñoù : (*) ⇔ 5 ⎡sin x + ( cos x − sin x ) ⎤ = 3 + 2 cos2 x − 1 ⎣ ⎦ 1⎞ ⎛ ⎜ do sin 2x ≠ − ⎟ 2⎠ ⎝ ⇔ 2 cos2 x − 5 cos x + 2 = 0

(

)

1 ⎡ cos x = 2 ⇔⎢ ⎢ ⎢cos x = 2 ( loaïi ) ⎣ π 3 1 ⇔ x = ± + k2π (nhaä n do sin 2x = ± ≠− ) 2 2 3 π 5π Do x ∈ ( 0, 2π ) neâ n x = ∨ x = 3 3
Baø i 57: (Ñeà thi tuyeån sinh Ñaï i hoï c khoái A, naê m 2005) Giaû i phöông trình: cos2 3x.cos 2x − cos2 x = 0 ( *) Ta coù : (*) ⇔

1 + cos 6x 1 + cos 2x .cos 2x − =0 2 2 ⇔ cos 6x.cos 2x − 1 = 0 (**) Caù c h 1: (**) ⇔ 4 cos3 2x − 3 cos 2x cos 2x − 1 = 0

(

)

⇔ 4 cos4 2x − 3 cos2 2x − 1 = 0 ⎡cos2 2x = 1 ⇔⎢ 2 ⎢cos 2x = − 1 ( voâ nghieäm ) ⎢ 4 ⎣ ⇔ sin 2x = 0

⇔ 2x = kπ ⇔ x =
Caù c h 2: (**) ⇔

1 ( cos 8x + cos 4x ) − 1 = 0 2 ⇔ cos 8x + cos 4x − 2 = 0
⇔ 2 cos2 4x + cos 4x − 3 = 0

kπ ( k ∈ Z) 2

⎡cos 4x = 1 ⇔⎢ ⎢cos 4x = − 3 ( loaïi ) 2 ⎣ kπ ⇔ 4x = k2π ⇔ x = ( k ∈ Z) 2 Caù c h 3: phöông trình löôï n g giaù c khoâ n g maã u möï c : ⎡cos 6x = cos 2x = 1 (**) ⇔ ⎢ ⎣cos 6x = cos 2x = −1

Caù c h 4: cos 8x + cos 4x − 2 = 0 ⇔ cos 8x + cos 4x = 2 ⇔ cos 8x = cos 4x = 1 ⇔ cos 4x = 1

Baø i 58: (Ñeà thi tuyeån sinh Ñaï i hoï c khoái D, naê m 2005) π⎞ π⎞ 3 ⎛ ⎛ Giaû i phöông trình: cos4 x + sin 4 x + cos ⎜ x − ⎟ sin ⎜ 3x − ⎟ − = 0 4⎠ 4⎠ 2 ⎝ ⎝

Ta coù : (*) ⎤ 3 1⎡ π⎞ ⎛ ⎢sin ⎜ 4x − 2 ⎟ + sin 2x ⎥ − 2 = 0 2⎣ ⎝ ⎠ ⎦ 1 1 3 ⇔ 1 − sin2 2x + [ − cos 4x + sin 2x ] − = 0 2 2 2 1 1 1 1 ⇔ − sin2 2x − 1 − 2 sin2 2x + sin 2x − = 0 2 2 2 2 2 ⇔ sin 2x + sin 2x − 2 = 0 ⎡sin 2x = 1 ⇔⎢ ⎣sin 2x = −2 ( loaïi ) π ⇔ 2x = + k2π, k ∈ 2 π ⇔ x = + kπ, k ∈ 4 ⇔ sin2 x + cos2 x

(

)

2

− 2 sin2 x cos2 x +

(

)

Baø i 59: (Ñeà th i tuyeån sinh Ñaï i ho ï c khoá i B, naê m 2004) Giaû i phöông trình: 5 sin x − 2 = 3 (1 − sinx ) tg 2 x Ñieà u kieä n : cos x ≠ 0 ⇔ sin x ≠ ±1 Khi ñoù: (*) ⇔ 5 sin x − 2 = 3 (1 − sin x )

( *)

⇔ 5sin x − 2 = 3 (1 − sin x )

3sin2 x ⇔ 5 sin x − 2 = 1 + sin x 2 ⇔ 2 sin x + 3sin x − 2 = 0 1 ⎡ ⎢sin x = 2 ( nhaän do sin x ≠ ±1) ⇔ ⎢ ⎢sin x = −2 ( voâ nghieäm ) ⎣

sin2 x 1 − sin2 x

sin2 x cos2 x

⇔x=

π 5π + k2π ∨ x = + k2π ( k ∈ Z) 6 6 1 1 = 2 cos 3x + ( *) sin x cos x

Baø i 60: Giaûi phöông trình: 2 sin 3x − Ñieà u kieä n : sin 2x ≠ 0

Luù c ñoù : (*) ⇔ 2 ( sin 3x − cos 3x ) =

1 1 + sin x cos x

1 1 ⇔ 2 ⎡3 ( sin x + cos x ) − 4 sin3 x + cos3 x ⎤ = + ⎣ ⎦ sin x cos x sin x + cos x ⇔ 2 ( sin x + cos x ) ⎡3 − 4 sin2 x − sin x cos x + cos2 x ⎤ = ⎣ ⎦ sin x cos x 1 ⎡ ⎤ ⇔ ( sin x + cos x ) ⎢ −2 + 8 sin x cos x − =0 sin x cos x ⎥ ⎣ ⎦ 2 ⎡ ⎤ ⇔ ( sin x + cos x ) ⎢4 sin 2x − − 2⎥ = 0 sin 2x ⎣ ⎦ ⎡ tgx = −1 ⎡sin x + cos x = 0 ⇔⎢ ⇔⎢ ( nhaän so vôùi ñieàu kieän ) 2 ⎢sin 2x = 1 ∨ sin 2x = −1 ⎣4 sin 2x − 2sin 2x − 2 = 0 2 ⎣

( (

)

)

π π π 7π + kπ ∨ 2x = + k2π ∨ 2x = − + k2π ∨ 2x = + k2π, k ∈ 4 2 6 6 π π 7π ⇔ x = ± + kπ ∨ x = − + kπ ∨ x = + kπ, k ∈ 4 12 12 ⇔x=−
Baø i 61: Giaûi phöông trình:
cos x 2 sin x + 3 2 − 2 cos2 x − 1

=1 1 + sin 2x π Ñieà u kieä n : sin 2x ≠ −1 ⇔ x ≠ − + mπ 4 Luù c ñoù : (*) ⇔ 2 sin x cos x + 3 2 cos x − 2 cos2 x − 1 = 1 + sin 2x ⇔ 2 cos2 x − 3 2 cos x + 2 = 0 2 ⇔ cos x = hay cos x = 2 ( voâ nghieäm ) 2 π ⎡ ⎢ x = 4 + k2π ⇔⎢ ⎢ x = − π + k '2π ( loaïi do ñieàu kieän ) ⎢ 4 ⎣ π ⇔ x = + k2π 4

(

)

( *)

Baø i 62: Giaûi phöông trình:

x 3x x 3x 1 cos x.cos .cos − sin x sin sin = ( *) 2 2 2 2 2

Ta coù : (*) ⇔

1 1 1 cos x ( cos 2x + cos x ) + sin x ( cos 2x − cos x ) = 2 2 2 2 ⇔ cos x.cos 2x + cos x + sin x cos 2x − sin x cos x = 1 ⇔ cos 2x ( cos x + sin x ) = 1 − cos2 x + sin x cos x
⇔ cos 2x ( cos x + sin x ) = sin x ( sin x + cos x )

⇔ ( cos x + sin x )( cos 2x − sin x ) = 0 ( * * )

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

(

)

⎡ cos x = − sin x ⇔⎢ 2 ⎣ 2 sin x + sin x − 1 = 0 π ⎡ ⎡ ⎢ x = − 4 + kπ ⎢ tgx = −1 ⎢ ⎢ π ⇔ ⎢sin x = −1 ⇔ ⎢ x = − + k2π ( k ∈ Z) ⎢ 2 ⎢ 1 ⎢ ⎢sin x = ⎢ x = π + k2π ∨ x = 5π + k2π 2 ⎣ ⎢ 6 6 ⎣ ⎛π ⎞ Caù c h khaù c: (**) ⇔ tgx = −1 ∨ cos 2x = sin x = cos ⎜ − x ⎟ ⎝2 ⎠
Baø i 63: Giaûi phöông trình: 4 cos3 x + 3 2 sin 2x = 8 cos x ( *) Ta coù : (*) ⇔ 4 cos3 x + 6 2 sin x cos x − 8 cos x = 0 ⇔ cos x 2 cos2 x + 3 2 sin x − 4 = 0

(

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

(

)

)

⇔x=

π 2 π + kπ ∨ sin x = = sin 4 2 2 π π 3π ⇔ x = + kπ ∨ x = + k2π ∨ x = + k2π ( k ∈ Z ) 2 4 4

Baø i 64 : Giaûi phöông trình: π⎞ π⎞ ⎛ ⎛ cos ⎜ 2x + ⎟ + cos ⎜ 2x − ⎟ + 4 sin x = 2 + 2 (1 − sin x ) ( *) 4⎠ 4⎠ ⎝ ⎝ (*) ⇔ 2 cos 2x.cos

⇔

π + 4 sin x = 2 + 2 (1 − sin x ) 4 2 1 − 2 sin2 x + 4 + 2 sin x − 2 − 2 = 0

(

⇔ 2 2 sin2 x − 4 + 2 sin x + 2 = 0

(

) (

)

)

⎡sin x = 2 ( loaïi ) ⇔ 2 sin x − 2 2 + 1 sin x + 2 = 0 ⇔ ⎢ ⎢sin x = 1 ⎢ 2 ⎣ π 5π ⇔ x = + k2π hay x = + k2π, k ∈ 6 6
2

(

)

Baø i 65 : Giaû i phöông trình : 3 cot g 2 x + 2 2 sin 2 x = 2 + 3 2 cos x ( * ) Ñieà u kieä n : sin x ≠ 0 ⇔ cos x ≠ ±1 Chia hai veá (*) cho sin 2 x ta ñöôï c : cos2 x cos x +2 2 = 2+3 2 vaø sin x ≠ 0 (*) ⇔ 3 4 sin x sin2 x cos x Ñaët t = ta ñöôï c phöông trình: sin 2 x 3t 2 − 2 + 3 2 t + 2 2 = 0

(

)

(

)

(

)

⇔t= 2∨t=
* Vôù i t =

2 cos x 2 ta coù : = 2 3 sin x 3 ⇔ 3 cos x = 2 1 − cos2 x

2 3

(

)

⇔ 2 cos2 x + 3 cos x − 2 = 0 ⎡cos x = −2 ( loaïi ) ⇔⎢ ⎢cos x = 1 ( nhaän do cos x ≠ ±1) ⎢ 2 ⎣ π ⇔ x = ± + k2π ( k ∈ Z ) 3 cos x * Vôù i t = 2 ta coù : = 2 sin2 x ⇔ cos x = 2 1 − cos2 x

(

)

⇔

2 cos2 x + cos x − 2 = 0

⎡cos x = − 2 ( loaïi ) ⎢ ⇔⎢ 2 ( nhaän do cos x ≠ ±1) ⎢cos x = 2 ⎣ π ⇔ x = ± + k2π, k ∈ 4

Baø i 66 : Giaûi phöông trình:

4 sin2 2x + 6 sin 2 x − 9 − 3 cos 2x = 0 ( *) cos x

Ñieà u kieä n : cos x ≠ 0 Luù c ñoù : (*) ⇔ 4 sin2 2x + 6 sin2 x − 9 − 3 cos 2x = 0
⇔ 4 1 − cos2 2x + 3 (1 − cos 2x ) − 9 − 3 cos 2x = 0 ⇔ 4 cos2 2x + 6 cos 2x + 2 = 0 ⇔ cos 2x = −1 ∨ cos 2x = − 1 2

(

)

⇔ 2 cos2 x − 1 = −1 ∨ 2 cos2 x − 1 = −

⎡cos x = 0 ( loaïi do ñieàu kieän ) ⇔⎢ ⎢cos x = ± 1 nhaän do cos x ≠ 0 ( ) ⎢ 2 ⎣ 2π π ⇔ x = ± + k2π ∨ x = ± + k2π ( k ∈ Z ) 3 3
Baø i 67: Cho f ( x ) = sin x +

1 2

1 2 sin 3x + sin 5x 3 5 Giaû i phöông trình: f ' ( x ) = 0
f '(x) = 0 ⇔ cos x + cos 3x + 2 cos 5x = 0

Ta coù :

⇔ ( cos x + cos 5x ) + ( cos 3x + cos 5x ) = 0 ⇔ 2 cos 3x cos 2x + 2 cos 4x cos x = 0 ⇔ 4 cos3 x − 3 cos x cos 2x + 2 cos2 2x − 1 cos x = 0
⇔ ⎡ 4 cos2 x − 3 cos 2x + 2 cos2 2x − 1⎤ cos x = 0 ⎣ ⎦ ⎡ ⎡ 2 (1 + cos 2x ) − 3⎤ cos 2x + 2 cos2 2x − 1 = 0 ⎦ ⇔ ⎢⎣ ⎢cos x = 0 ⎣ ⎡4 cos2 2x − cos 2x − 1 = 0 ⇔⎢ ⎣cos x = 0 1 ± 17 ∨ cos x = 0 8 1 + 17 1 − 17 ⇔ cos 2x = = cos α ∨ cos 2x = = cos β ∨ cos x = 0 8 8 α β π ⇔ x = ± + kπ ∨ x = ± + kπ ∨ x = + kπ ( k ∈ Z ) 2 2 2 ⇔ cos 2x =

(

(

)

)

(

)

Baø i 68: Giaûi phöông trình: sin8 x + cos8 x = Ta coù :

17 cos2 2x ( *) 16

sin 8 x + cos8 x = sin4 x + cos4 x

(

)

2

− 2 sin 4 x cos4 x
2

= ⎡ sin 2 x + cos2 x ⎢ ⎣
2

(

)

1 − 2 sin 2 x cos2 x ⎤ − sin4 2x ⎥ 8 ⎦

2

Do ñoù :

1 1 ⎛ ⎞ = ⎜ 1 − sin2 2x ⎟ − sin 4 2x 2 8 ⎝ ⎠ 1 = 1 − sin2 2x + sin4 2x 8

( *) ⇔ 16 ⎛ 1 − sin2 2x + ⎜
⎝

1 ⎞ sin4 2x ⎟ = 17 1 − sin2 2x 8 ⎠

(

)

⇔ 2 sin4 2x + sin2 2x − 1 = 0 ⎡sin2 2x = −1 ( loaïi ) 1 1 ⇔⎢ ⇔ (1 − cos 4x ) = 1 ⎢sin2 2x = 2 2 ⎢ 2 ⎣ π ⇔ cos 4x = 0 ⇔ x = ( 2k + 1) , ( k ∈ Z ) 8
Baø i 69 : Giaûi phöông trình: sin

5x x = 5 cos3 x.sin ( *) 2 2

x = 0 ⇔ x = π + k2π ⇔ cos x = −1 2 Thay vaø o (*) ta ñöôï c : ⎛ 5π ⎞ ⎛π ⎞ sin ⎜ + 5kπ ⎟ = − 5. sin ⎜ + kπ ⎟ , khoâ n g thoû a ∀k ⎝ 2 ⎠ ⎝2 ⎠ x Do cos khoâ n g laø nghieä m cuû a (*) neâ n : 2 5x x x x x ( *) ⇔ sin . cos = 5 cos2 x. sin cos vaø cos ≠ 0 2 2 2 2 2 1 5 x ⇔ ( sin 3x + sin 2x ) = cos3 x.sin x vaø cos ≠ 0 2 2 2
Nhaän xeù t thaáy : cos

⇔ 3sin x − 4 sin3 x + 2 sin x cos x = 5 cos3 x.sin x vaø cos

x ⎧ ⎪cos ≠ 0 2 ⇔⎨ ⎪3 − 4 sin2 x + 2 cos x = 5 cos3 x ∨ sin x = 0 ⎩

x ≠0 2

⇔

⇔

⇔

⇔

x ⎧ cos ≠ 0 ⎪ ⎪ 2 ⎨ ⎪5 cos3 x − 4 cos2 x − 2 cos x + 1 = 0 ∨ sin x = 0 ⎪ 2 ⎩ ⎧cos x ≠ −1 ⎪ ⎨ x 2 ⎪( cos x − 1) 5 cos x + cos x − 1 = 0 ∨ sin 2 = 0 ⎩ ⎧cos x ≠ −1 ⎪ ⎪⎡ ⎪ ⎢cos x = 1 ⎪⎢ −1 + 21 ⎨⎢ = cos α ⎪ ⎢cos x = 10 ⎪⎢ −1 − 21 ⎪⎢ = cos β ⎪ ⎣cos x = ⎢ 10 ⎩ x = k2π hay x = ±α + k2π hay x = ±β + k2π, ( k ∈ Z )

(

)

Baø i 70: Giaûi phöông trình: sin 2x ( cot gx + tg2x ) = 4 cos2 x ( *) Ñ ieà u kieä n : cos 2x ≠ 0 vaø sin x ≠ 0 ⇔ cos 2x ≠ 0 ∧ cos 2x ≠ 1 cos x sin 2x + Ta coù : cot gx + tg2x = 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 ⎠ 2 cos x ⇔ = 2 cos2 x cos 2x ⇔ ( cos 2x + 1) = 2 cos 2x ( cos 2x + 1)

⇔ ( cos 2x + 1) = 0 hay 1 = 2 cos 2x ⇔ cos 2x = −1 ∨ cos 2x = 1 ( nhaän do cos 2x ≠ 0 vaø cos 2x ≠ 1) 2 π ⇔ 2x = π + k2π ∨ 2x = ± + k2π, k ∈ 3 π π ⇔ x = + kπ ∨ x = ± + kπ, k ∈ 2 6

Baø i 71 : Giaûi phöông trình: 2 cos2

6x 8x + 1 = 3 cos ( *) 5 5

12x ⎞ ⎛ ⎛ ⎞ 2 4x Ta coù : (*) ⇔ ⎜ 1 + cos − 1⎟ ⎟ + 1 = 3 ⎜ 2 cos 5 ⎠ 5 ⎝ ⎝ ⎠ 4x 4x 4x ⎞ ⎛ ⇔ 2 + 4 cos3 − 3 cos = 3 ⎜ 2 cos2 − 1⎟ 5 5 5 ⎝ ⎠ 4 Ñaë t t = cos x ( ñieàu kieän t ≤ 1) 5 Ta coù phöông trình : 4t 3 − 3t + 2 = 6t 2 − 3
⇔ 4t 3 − 6t 2 − 3t + 5 = 0 ⇔ ( t − 1) ( 4t 2 − 2t − 5 ) = 0 ⇔ t = 1∨ t = 1 − 21 1 + 21 ∨t = ( loïai ) 4 4

Vaä y

4x 4x =1⇔ = 2kπ 5 5 5kπ ⇔x= ( k ∈ Z) 2 4x 1 − 21 • cos = = cos α ( vôùi 0 < α < 2 π ) 5 4 4x ⇔ = ±α + l 2 π 5 5α l 5π ⇔x=± + ,(l ∈ Z) 4 2
• cos

π⎞ ⎛ Baø i 72 : Giaûi phöông trình tg3 ⎜ x − ⎟ = tgx − 1 ( *) 4⎠ ⎝ π π Ñaë t t = x − ⇔ x = + t 4 4 1 + tgt ⎛π ⎞ (*) thaø n h : tg3 t = tg ⎜ + t ⎟ − 1 = − 1 vôùi cos t ≠ 0 ∧ tgt ≠ 1 1 − tgt ⎝4 ⎠ 2tgt ⇔ tg3 t = 1 − tgt ⇔ tg3 t − tg 4 t = 2tgt ⇔ tgt ( tg3 t − tg 2 t + 2 ) = 0 ⇔ tgt ( tgt + 1) ( tg 2 t − 2tgt + 2 ) = 0 ⇔ tgt = 0 ∨ tgt = −1( nhaän so ñieàu kieän ) ⇔ t = kπ ∨ t = − π + kπ, k ∈¢ 4

Vaä y (*)

⇔x=

π + kπ hay x = kπ, k ∈¢ 4

sin 4 2x + cos4 2x = cos4 4x (*) Baø i 73 : Giaûi phöông trình ⎛π ⎞ ⎛π ⎞ tg ⎜ − x ⎟ tg ⎜ + x ⎟ ⎝4 ⎠ ⎝4 ⎠
Ñieà u kieä n ⎧ ⎛π ⎧ ⎛π ⎞ ⎛π ⎞ ⎞ ⎪sin ⎜ 4 − x ⎟ cos ⎜ 4 − x ⎟ ≠ 0 ⎪sin ⎜ 2 − 2x ⎟ ≠ 0 ⎪ ⎝ ⎠ ⎝ ⎠ ⎪ ⎝ ⎠ ⇔⎨ ⎨ ⎪sin ⎛ π + x ⎞ cos ⎛ π + x ⎞ ≠ 0 ⎪sin ⎛ π + 2x ⎞ ≠ 0 ⎜4 ⎟ ⎜4 ⎟ ⎟ ⎪ ⎝ ⎪ ⎜2 ⎠ ⎝ ⎠ ⎠ ⎩ ⎩ ⎝ ⇔ cos 2x ≠ 0 ⇔ sin 2x ≠ ±1 Do : ⎛π ⎞ ⎛π ⎞ 1 − tgx 1 + tgx tg ⎜ − x ⎟ tg ⎜ + x ⎟ = . =1 ⎝4 ⎠ ⎝4 ⎠ 1 + tgx 1 − tgx Khi cos2x ≠ 0 thì : (*) ⇔ sin 4 2x + cos4 2x = cos4 4x

⇔ 1 − 2 sin 2 2x cos2 2x = cos4 4x 1 ⇔ 1 − sin 2 4x = cos4 4x 2 1 ⇔ 1 − (1 − cos2 4x ) = cos4 4x 2 ⇔ 2 cos4 4x − cos2 4x − 1 = 0 ⎡ cos2 4x = 1 ⇔⎢ 2 ⇔ 1 − sin 2 4x = 1 ⎢ cos 4x = − 1 ( voâ nghieäm ) ⎢ ⎣ 2 ⇔ sin 4x = 0 ⇔ 2 sin 2x cos 2x = 0 ⇔ sin 2x = 0 ( do cos 2x ≠ 0 ) π ⇔ 2x = kπ, k ∈¢ ⇔ x = k , k ∈¢ 2 1 2 − 2 (1 + cot g2x cot gx ) = 0 ( *) Baø i 74 :Giaû i phöông trình: 48 − 4 cos x sin x Ñieà u kieä n : sin 2x ≠ 0 Ta coù :

cos 2x cos x . sin 2x sin x sin 2x sin x + cos 2x cos x = sin x sin 2x cos x 1 = = ( do cos x ≠ 0 ) 2 2 sin x cos x 2 sin 2 x 1 1 − 4 =0 Luù c ñoù (*) ⇔ 48 − 4 cos x sin x 1 1 sin 4 x + cos4 x ⇔ 48 = + 4 = cos4 x sin x sin 4 x cos4 x ⇔ 48sin 4 x cos4 x = sin 4 x + cos4 x 1 + cot g2x cot gx = 1 +

⇔ 3sin 4 2x = 1 − 2 sin 2 x cos2 x 1 ⇔ 3sin 4 2x + sin 2 2x − 1 = 0 2 2 ⎡ 2 ⎢sin x = − 3 ( loïai ) ⇔⎢ ⎢sin 2 x = 1 ( nhaän do ≠ 0 ) ⎢ 2 ⎣

1 1 (1 − cos 4x ) = 2 2 ⇔ cos 4x = 0 π ⇔ 4x = + kπ 2 π kπ ⇔ x = + ( k ∈ Z) 8 4
⇔
Baø i 75 : Giaû i phöông trình

5 sin 8 x + cos8 x = 2 sin10 x + cos10 x + cos 2x ( *) 4

(

)

Ta coù : (*)

⇔ sin8 x − 2 sin10 x + cos8 x − 2 cos10 x =

5 cos 2x 4 5 ⇔ sin 8 x (1 − 2 sin 2 x ) − cos8 x ( −1 + 2 cos2 x ) = cos 2x 4 5 ⇔ sin 8 x.cos 2x − cos8 x cos 2x = cos 2x 4 8 8 ⇔ 4 cos 2x ( sin x − cos x ) = 5 cos 2x

(

) (

)

⇔ cos 2x = 0 hay 4 ( sin 8 x − cos8 x ) = 5 ⇔ cos 2x = 0 hay 4 ( sin 4 x − cos4 x )( sin 4 x + cos4 x ) = 5 ⎛ 1 ⎞ ⇔ cos 2x = 0 hay 4 ⎜ 1 − sin 2 2x ⎟ = 5 ⎝ 2 ⎠ 2 ⇔ cos 2x = 0 hay − 2 sin 2x = 1(Voâ nghieäm ) π ⇔ 2x = + kπ, k ∈¢ 2 π kπ ⇔x= + , k ∈¢ 4 2 Caù c h khaù c: Ta coù 4 ( sin 8 x − cos8 x ) = 5 voâ nghieä m

Vì

( sin

8

x − cos8 x ) ≤ 1, ∀ x neâ n 4 ( sin 8 x − cos8 x ) ≤ 4 < 5, ∀x

Ghi chuù : Khi gaë p phöông trình löôï n g giaùc daï n g R(tgx, cotgx, sin2x, cos2x, tg2x) vôù i R haø m höõ u tyû thì ñaë t t = tgx 2t 2t 1 − t2 , sin 2x = , cos 2x = Luù c ñoù tg2x = 1 − t2 1 + t2 1 + t2 Baø i 76 : (Ñeå thi tuyeån sinh Ñaïi hoï c khoái A, naêm 2003) Giaû i phöông trình cos 2x 1 cot gx − 1 = + sin2 x − sin 2x ( *) 1 + tgx 2

Ñieà u kieä n : sin 2x ≠ 0 vaø tgx ≠ −1 Ñaët t = tgx thì (*) thaø nh : 1 − t2 2 1 1 + t 2 + 1 ⎡1 − 1 − t ⎤ − 1 . 2t −1 = ⎢ ⎥ t 1+t 2⎣ 1 + t2 ⎦ 2 1 + t2
⇔

1−t 1 − t 1 2t 2 t = + . − ( do t ≠ −1) 2 2 2 1+t 1 + t2 t 1+t
2

1 − t t 2 − 2t + 1 (1 − t ) ⇔ = = t 1 + t2 1 + t2
⇔ ( 1 − t ) (1 + t 2 ) = ( 1 − t ) t
2

⎡ t = 1 ( nhaän do t ≠ −1) ⎡1 − t = 0 ⇔⎢ ⇔⎢ 2 2 ⎣1 + t = (1 − t ) t ⎢2t − t + 1 = 0 ( voâ nghieäm ) ⎣ π Vaä y (*) ⇔ tgx = 1 ⇔ x = + kπ ( nhaän do sin 2x = 1 ≠ 0) 4

Baø i 77 : Giaûi phöông trình: sin 2x + 2tgx = 3 ( * )

Ñieà u kieä n : cos x ≠ 0 Ñ aët t = tgx thì (*) thaøn h :

2t + 2t = 3 1 + t2 ⇔ 2t + ( 2t − 3) (1 + t 2 ) = 0
⇔ 2t 3 − 3t 2 + 4t − 3 = 0 ⇔ ( t − 1) ( 2t 2 − t + 3) = 0 ⎡t = 1 ⇔⎢ 2 ⎣2t − t + 3 = 0 ( voâ nghieäm ) π Vaäy (*) ⇔ tgx = 1 ⇔ x = + kπ ( k ∈ Z ) 4

Baø i 78 : Giaû i phöông trình

cot gx − tgx + 4 sin 2x =
Ñieà u kieä n : sin 2x ≠ 0

2 ( *) sin 2x

2t do sin 2x ≠ 0 neân t ≠ 0 1 + t2 1 8t 1 + t2 1 = = +t (*) thaø n h : − t + t 1 + t2 t t 8t ⇔ = 2t 1 + t2 4 ⇔ = 1 ( do t ≠ 0 ) 1 + t2 ⇔ t 2 = 3 ⇔ t = ± 3 ( nhaän do t ≠ 0 )
Ñaë t t = tgx thì : sin 2x =
Vaäy (*) ⎛ π⎞ ⇔ tgx = tg ⎜ ± ⎟ ⎝ 3⎠ π ⇔ x = ± + kπ, k ∈ 3

Baø i 79 : Giaû i phöông trình (1 − tgx )(1 + sin 2x ) = 1 + tgx ( * )

Ñieà u kieä n : cos x ≠ 0 Ñaët = tgx thì (*) thaø nh : 2t ⎞ (1 − t ) ⎛ 1 + ⎜ ⎟ =1+t 1 + t2 ⎠ ⎝ 2 ( t + 1) = 1 + t ⇔ (1 − t ) 1 + t2 ⎡ t = −1 ⎡ t = −1 ⇔ ⎢ (1 − t )(1 + t ) ⇔ ⎢ 2 2 ⎢ =1 ⎣1 − t = 1 + t 2 ⎢ 1+t ⎣ ⇔ t = −1 ∨ t = 0

⎡ tgx = −1 π Do ñoù (*) ⇔ ⎢ ⇔ x = − + kπ hay x = kπ, k ∈ 4 ⎣ tgx = 0
Baø i 80 : Cho phöông trình cos 2x − ( 2m + 1) cos x + m + 1 = 0 ( * )

a/ Giaû i phöông trình khi m =

⎛ π 3π ⎞ b/ Tìm m ñeå (*) coù nghieä m treâ n ⎜ , ⎟ ⎝2 2 ⎠ 2 Ta coù (*) 2 cos x − ( 2m + 1) cos x + m = 0

3 2

⎧t = cos x ([ t ] ≤ 1) ⎪ ⇔⎨ 2 ⎪2t − ( 2m + 1) t + m = 0 ⎩ ⎧ t = cos x ([ t ] ≤ 1) ⎪ ⇔⎨ 1 ⎪t = ∨ t = m ⎩ 2 3 a/ Khi m = , phöông trình thaønh 2 1 3 cos x = ∨ cos x = ( loaïi ) 2 2 π ⇔ x = ± + k2π ( k ∈ Z ) 3 ⎛ π 3π ⎞ b/ Khi x ∈ ⎜ , ⎟ thì cos x = t ∈ [−1, 0) ⎝2 2 ⎠ 1 Do t = ∉ [ −1, 0] neân 2 π 3π ( *) coù nghieäm treân ⎛ , ⎞ ⇔ m ∈ ⎡ −1, 0) ⎜ ⎟ ⎣ ⎝2 2 ⎠
Baø i 81 : Cho phöông trình ( cos x + 1)( cos 2x − m cos x ) = m sin 2 x ( *)

a/ Giaû i (*) khi m= -2

Ta coù (*) ⇔ ( cos x + 1) ( 2 cos2 x − 1 − m cos x ) = m (1 − cos2 x ) ⇔ ( cos x + 1) ⎡2 cos2 x − 1 − m cos x − m (1 − cos x ) ⎤ = 0 ⎣ ⎦ ⇔ ( cos x + 1) ( 2 cos2 x − 1 − m ) = 0

⎡ 2π ⎤ b/ Tìm m sao cho (*) coù ñuù n g hai nghieä m treâ n ⎢0, ⎥ ⎣ 3⎦

a/ Khi m = -2 thì (*) thaø nh :

( cos x + 1) ( 2 cos2 x + 1) = 0
⇔ cosx = -1 ⇔ x = π + k2π ( k ∈ Z ) ⎡ 2π ⎤ ⎡ 1 ⎤ b / Khi x ∈ ⎢ 0, ⎥ thì cos x = t ∈ ⎢ − ,1⎥ ⎣ 3⎦ ⎣ 2 ⎦ ⎡ 1 ⎤ Nhaä n xeù t raè n g vôù i moãi t treâ n ⎢ − ,1⎥ ta chæ tìm ñöôï c duy nhaá t moä t x treâ n ⎣ 2 ⎦ ⎡ 2π ⎤ ⎢0, ⎥ ⎣ 3⎦ ⎡ 1 ⎤ Yeâ u caà u baø i toaù n ⇔ 2t 2 − 1 − m = 0 coù ñu ù n g hai n ghieä m treâ n ⎢ − ,1⎥ ⎣ 2 ⎦ Xeù t y = 2t 2 − 1 ( P ) vaø y = m ( d )

Ta coù y’ = 4t

⎡ 2π ⎤ Vaä y (*) coù ñuù ng hai nghieä m treâ n ⎢0, ⎥ ⎣ 3⎦

⎡ 1 ⎤ ⇔ (d) caé t (P) taï i hai ñieå m phaân bieä t treâ n ⎢ − ,1⎥ ⎣ 2 ⎦ 1 ⇔ −1 < m ≤ 2

Baø i 82 : Cho phöông trình (1 − a ) tg 2 x −

a/ Giaû i (1) khi a =

1 2

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

⎛ π⎞ b/ Tìm a ñeå (1) coù nhieà u hôn moä t nghieä m treâ n ⎜ 0, ⎟ ⎝ 2⎠ π Ñieà u kieä n : cos x ≠ 0 ⇔ x ≠ + kπ 2

(1) ⇔ (1 − a ) sin2 x − 2 cos x + (1 + 3a ) cos2 x = 0 ⇔ (1 − a ) (1 − cos2 x ) − 2 cos x + (1 + 3a ) cos2 x = 0
⇔ 4a cos2 x − 2 cos x + 1 − a = 0 ⇔ a ( 4 cos2 x − 1) − ( 2 cos x − 1) = 0 ⇔ ( 2 cos x − 1) ⎡a ( 2 cos x + 1) − 1⎤ = 0 ⎣ ⎦

1 1⎞ ⎛ thì (1) thaø n h : ( 2 cos x − 1) ⎜ cos x − ⎟ = 0 2 2⎠ ⎝ 1 π ⇔ cos x = = cos ( nhaän do cos x ≠ 0 ) 2 3 π ⇔ x = ± + k2π ( k ∈ Z ) 3 ⎛ π⎞ b/ Khi x ∈ ⎜ 0, ⎟ thì cos x = t ∈ ( 0,1) ⎝ 2⎠ 1 ⎡ cos x = t = ∈ ( 0,1) 2 Ta coù : (1) ⇔ ⎢ ⎢ ⎢2a cos x = 1 − a ( 2 ) ⎣
a/ Khi a =
⎧ ⎪a ≠ 0 ⎪ 1−a ⎪ ⎧1 ⎫ Yeâ u caà u baø i toaù n ⇔ (2) coù nghieä m treâ n ( 0,1) \ ⎨ ⎬ ⇔ ⎨0 < <1 2a ⎩2⎭ ⎪ ⎪1 − a 1 ⎪ 2a ≠ 2 ⎩ ⎧a ≠ 0 ⎧ ⎪1 − a ⎪0 < a < 1 ⎧1 ⎪ >0 ⎪ ⎪3 < a < 1 1 ⎪ 2a ⎪ ⎪ ⇔⎨ ⇔ ⎨a < 0 ∨ a > ⇔ ⎨ 1 − 3a 3 ⎪ ⎪ ⎪a ≠ 1 <0 ⎪ 1 ⎪ 2a ⎪ 2 ⎩ ⎪2 (1 − a ) ≠ 2a ⎪a ≠ 2 ⎩ ⎩

Caù c h khaù c : daë t u =

(1 − a ) ( u 2 − 1 ) − 2u + 1 + 3a = 0 ⇔ (1 − a ) u 2
⇔ ( u − 2 ) [ (1 − a)u − 2a ] = 0

1 , ñieà u kieä n u ≥1 ; pt thaø n h cos x

− 2u + 4a = 0

Baø i 83 : Cho phöông trình : cos 4x + 6 sin x cos x = m (1)

a/ Giaû i (1) khi m = 1
⎡ π⎤ b/ Tìm m ñeå (1) coù hai nghieä m phaân bieä t treâ n ⎢ 0, ⎥ ⎣ 4⎦ 2 Ta coù : (1) ⇔ 1 − 2 sin 2x + 3 sin 2x = m

⎧t = sin 2x ( t ≤ 1) ⎪ ⇔⎨ 2 ⎪2t − 3t + m − 1 = 0 ( 2 ) ⎩
a/ Khi m = 1 thì (1) thaø nh ⎧t = sin 2x ( t ≤ 1) ⎧ t = sin 2x ( t ≤ 1) ⎪ ⎪ ⇔⎨ ⎨ 2 3 ⎪2t − 3t = 0 ⎪t = 0 ∨ t = ( loaïi ) ⎩ 2 ⎩ kπ ⇔ sin 2x = 0 ⇔ x = 2 ⎡ π⎤ b/ Khi x ∈ ⎢0, ⎥ thì sin 2x = t ∈ [ 0,1] ⎣ 4⎦ Nhaän thaáy raè n g moãi t tìm ñöôïc treâ n [ 0,1] ta chæ tìm ñöôïc duy nhaá t moä t
⎡ π⎤ x ∈ ⎢ 0, ⎥ ⎣ 4⎦ Ta coù : (2) ⇔ −2t 2 + 3t + 1 = m Xeù t y = −2t 2 + 3t + 1 treân [ 0,1]

Thì y ' = −4t + 3

Yeâ u caà u baø i toaù n ⇔ (d) y = m caé t taï i hai ñieå m phaâ n bieä t treâ n [ 0,1]

⇔2 ≤ m <

Caù c h khaù c :ñaët f (x) = 2t 2 − 3t + m − 1 . Vì a = 2 > 0, neâ n ta coù
⎧Δ =17 − 8m > 0 ⎪ f (0) = m −1≥ 0 ⎪ 17 ⎪ Yeâ u caà u baø i toaù n ⇔ ⎨ f (1) = m − 2 ≥ 0 ⇔ 2 ≤ m < 8 ⎪ S 3 ⎪ 0 ≤ = ≤1 ⎪ 2 4 ⎩

17 8

Baø i 84 : Cho phöông trình 4 cos5 x.sin x − 4 sin 5 x cos x = sin 2 4x + m (1 )

a/ Bieát raè ng x = π laø nghieäm cuûa (1). Haõ y giaûi (1) trong tröôøn g hôï p ñoù . π b/ Cho bieá t x = − laø moä t nghieä m cuû a (1). Haõ y tìm taá t caû nghieä m cuû a (1) thoû a 8 4 2 x − 3x + 2 < 0

(1) ⇔ 4 sin x cos x ( cos4 x − sin 4 x ) = sin2 4x + m

⇔ 2 sin 2x ( cos2 x − sin2 x )( cos2 x + sin 2 x ) = sin 2 4x + m ⇔ 2 sin 2x.cos 2x = sin 2 4x + m ⇔ sin 2 4x − sin 4x + m = 0

(1)

a/ x = π laø nghieä m cuû a (1) ⇒ sin2 4π − sin 4π + m = 0 ⇒m = 0 Luù c ñoù (1) ⇔ sin 4x (1 − sin 4x ) = 0
⇔ sin 4x = 0 ∨ sin 4x = 1 π + k2π 2 kπ π kπ ⇔x = ∨x= + ( k ∈ Z) 4 8 2 ⎧t = x2 ≥ 0 ⎧t = x2 ≥ 0 ⎪ ⇔⎨ b/ x 4 − 3x 2 + 2 < 0 ⇔ ⎨ 2 ⎪t − 3t + 2 < 0 ⎩1 < t < 2 ⎩ ⇔ 1 < x2 < 2 ⇔ 1 < x < 2 ⇔ 4x = kπ ∨ 4x =

⇔ − 2 < x < −1 ∨ 1 < x < 2 ( *) π ⎛ π⎞ thì sin 4x = sin ⎜ − ⎟ = −1 8 ⎝ 2⎠ π x = − laø nghieäm cuûa (1) ⇒ 1 + 1 + m = 0 8 ⇒ m = −2 x=−

⎧t = sin 4x ( vôùi t ≤ 1) ⎪ ⇔⎨ ⎪t = −1 ∨ t = 2 ( loaïi ) ⎩ ⇔ sin 4x = −1 π ⇔ 4x = − + k2π 2 π kπ ⇔x = − + 8 2 Keá t hôï p vôù i ñi eà u kieä n (*) suy ra k = 1 π π 3π Vaä y (1) coù n ghieä m x = − + = thoû a x4 − 3x2 + 2 < 0 8 2 8 Baø i 85 : Tìm a ñeå hai phöông trình sau töông ñöông 2 cos x.cos 2x = 1 + cos 2x + cos 3x (1 ) 4 cos2 x − cos 3x = a cos x + ( 4 − a )(1 + cos 2x )

Luù c ñoù (1) thaø nh : sin2 4x − sin 4x − 2 = 0 ⎧t = sin 4x ( vôùi t ≤ 1) ⎪ ⇔⎨ 2 ⎪t − t − 2 = 0 ⎩

( 2)

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

(

)

(

)

⇔ 4 cos3 x + ( 4 − 2a ) cos2 x ( a − 3) cos x = 0 ⎡cos x = 0 ⇔⎢ 2 ⎢4 cos x + 2 ( 2 − a ) cos x + a − 3 = 0 ⎣ 1⎞ ⎛ ⇔ cos x = 0 hay ⎜ cos x − ⎟ [ 2 cos x + 3 − a ] = 0 2⎠ ⎝ 1 a−3 ⇔ cos x = 0 ∨ cos x = ∨ cos x = 2 2

Vaä y yeâ u caà u baø i toaù n ⎡a − 3 ⎢ 2 =0 ⎢ ⎢a − 3 = 1 ⇔ ⇔ ⎢ 2 2 ⎢a − 3 a−3 ⎢ < −1 ∨ >1 ⎢ 2 ⎣ 2

⎡a = 3 ⎢a = 4 ⎢ ⎢ ⎣a < 1 ∨ a > 5

Baø i 86 : Cho phöông trình : cos4x = cos 2 3x + asin 2 x (*) a/ Giaû i phöông trì nh khi a = 1 ⎛ π ⎞ b/ Tìm a ñeå (*) coù nghieä m treâ n ⎜ 0, ⎟ ⎝ 12 ⎠ 1 a Ta coù : ( *) ⇔ cos 4x = (1 + cos 6x ) + (1 − cos 2x ) 2 2 2 3 ⇔ 2 2 cos 2x − 1 = 1 + 4 cos 2x − 3 cos 2x + a (1 − cos 2x )

(

)

⎧t = cos 2x ( t ≤ 1) ⎪ ⇔⎨ 2 3 ⎪2 2t − 1 = 1 + 4t − 3t + a (1 − t ) ⎩ ⎧t = cos 2x ( t ≤ 1) ⎪ ⇔⎨ 3 2 ⎪−4t + 4t + 3t − 3 = a (1 − t ) ⎩ ⎧1 = cos 2x ( t ≤ 1) ⎪ ⇔⎨ 2 ⎪( t − 1) −4t + 3 = a (1 − t ) ( * *) ⎩ a/ Khi a = 1 thì (*) thaø nh :

(

)

(

)

⎧ ⎧ ⎪t = cos 2x ( t ≤ 1) ⎪t = cos 2x ⇔⎨ ⎨ 2 ⎪t = ±1 ⎪( t − 1) −4t + 4 = 0 ⎩ ⎩ ⇔ cos 2x = ±1 ⇔ cos2 2x = 1

(

)

(t

≤ 1)

kπ , ( k ∈ Z) 2 ⎛ 3 ⎞ ⎛ π ⎞ ⎛ π⎞ b/ Ta coù : x ∈ ⎜ 0, ⎟ ⇔ 2x ∈ ⎜ 0, ⎟ .Vaäy cos 2x = t ∈ ⎜ ⎜ 2 ,1 ⎟ ⎟ ⎝ 12 ⎠ ⎝ 6⎠ ⎝ ⎠ ⇔ sin 2x = 0 ⇔ 2x = kπ ⇔ x =
Vaäy (**) ⇔ ( t-1) −4t 2 + 3 = a (1 − t ) ⇔ 4t 2 − 3 = a ( do t ≠ 1) ⎛ 3 ⎞ X eù t y = 4t 2 − 3 ( P ) treân ⎜ ⎜ 2 ,1 ⎟ ⎟ ⎝ ⎠ ⎛ 3 ⎞ ⇒ y ' = 8t > 0 ∀t ∈ ⎜ ⎜ 2 ,1 ⎟ ⎟ ⎝ ⎠ ⎛ 3 ⎞ ⎛ π⎞ ,1 ⎟ Do ñ o ù (*) coù nghieä m treâ n ⎜ 0, ⎟ ⇔ ( d ) : y = a caét ( P ) treân ⎜ ⎜ 2 ⎟ ⎝ 2⎠ ⎝ ⎠ ⎛ 3⎞ ⇔ y⎜ ⎜ 2 ⎟ < a < y (1 ) ⎟ ⎝ ⎠ ⇔ 0<a <1

(

)

1.

Gia û i ca ù c phöông trình sau : a/ sin4x = tgx π⎞ π⎞ 9 ⎛ ⎛ b/ sin4 x + sin4 x ⎜ x + ⎟ + sin4 ⎜ x − ⎟ = 4⎠ 4⎠ 8 ⎝ ⎝ c/ tgx + cot gx = 4
1 − sin 2x 4 e/ 4 cos x + 3 2 sin 2x = 8 cos x 1 1 2 + = f/ cos x sin 2x sin 4x π⎞ ⎛ g/ sin 2x + 2 sin ⎜ x − ⎟ = 1 4⎠ ⎝

BAØI TAÄP

d/

sin x 3 2 − 2 cos x − 2 sin 2 x − 1

(

)

=1

π⎞ π⎞ ⎛ ⎛ 2 ( 2 sin x − 1) = 4 ( sin x − 1) − cos ⎜ 2x + ⎟ − sin ⎜ 2x + ⎟ 4⎠ 4⎠ ⎝ ⎝ 4x = cos2 x k/ cos 3 x l/ tg .cos x + sin 2x = 0 2
h/

m/ 1 + 3tgx = 2sin 2x n/ cot gx = tgx + 2tg2x 3x 4x + 1 = 3 cos p/ 2 cos2 5 5 2 q/ 3 cos 4x − 2 cos 3x = 1 3x + 1 = 3 cos 2x r/ 2 cos2 2 x s/ cos x + tg = 1 2 t/ 3tg2x − 4tg3x = tg 2 3x.tg2x u/ cos x.cos 4x + cos 2x.cos 3x + cos2 4x = v/ cos2 x + cos2 2x + cos2 3x + cos2 4x = w/ sin 4x = tgx

3 2

3 2

2.

13 cos2 2x 8 ⎛ 3π x ⎞ 1 ⎛ π 3x ⎞ y/ sin ⎜ − ⎟ = sin ⎜ + ⎟ ⎝ 10 2 ⎠ 2 ⎝ 10 2 ⎠ sin6 x + cos6 x = a sin 2x (1) a/ Giaû i phöông trình khi a = 1.
x/ cos6 x + sin6 x = b/ Tìm a ñeå (1) coù nghieä m (ÑS : a ≥

3.

Cho phöông trình cos6 x + sin6 x = 2mtg2x cos2 x − sin2 x

1 ) 4

(1 )
1 8

a/ Giaû i phöông trình khi m =

b/ Tìm m sao cho (1) coù nghieä m 4. Tìm m ñeå phöông trình sin 4x = mtgx coù nghieäm x ≠ kπ

(ÑS : m ≥

1 ) 8

1 ⎛ ⎞ ⎜ ÑS : − < m < 4 ⎟ 2 ⎝ ⎠
5. Tìm m ñeå phöông trình : cos 3x − cos 2x + m cos x − 1 = 0 ⎛ π ⎞ coù ñuù n g 7 nghieä m treâ n ⎜ − , 2π ⎟ ( ÑS :1 < m < 3) ⎝ 2 ⎠ Tìm m ñeå phöông trình : 4 sin 4 x + cos4 x − 4 sin 6 x + cos6 x − sin 2 4x = m coù nghieä m

6.

(

) (

)

1 ⎛ ⎞ ⎜ ÑS : − ≤ m ≤ 1 ⎟ 8 ⎝ ⎠

7.

Cho phöông trình : 6 sin2 x − sin2 x = m cos2 2x a/ Giaû i phöông trình khi m = 3 b/ Tìm m ñeå (1) coù nghieä m

(1)

( ÑS : m ≥ 0 )

8.

9.

Tìm m ñeå phöông trình : ( 2m + 1) sin2 x = 0 m sin4 x + cos 4x + sin 4x − 4 4 π π⎞ ⎛ coù hai nghieä m phaâ n bieä t treâ n ⎜ , ⎟ ⎝4 2⎠ 1⎞ ⎛ ⎜ ÑS : 2 5 − 4 < m < ⎟ 2⎠ ⎝ Tìm m ñeå phöông trình : sin6 x + cos6 x = m sin 4 x + cos4 x coù nghieä m

(

)

1 ⎛ ⎞ ⎜ ÑS : ≤ m ≤ 1 ⎟ 2 ⎝ ⎠

10.

Cho phöông trình : cos 4x = cos2 3x + a sin2 x
⎛ π⎞ Tìm a ñeå phöông trình coù nghieä m x ∈ ⎜ 0, ⎟ ⎝ 2⎠

( ÑS : 0 < a < 1)
Th.S Phạm Hồng Danh

TT luyện thi đại học CLC Vĩnh Viễn


				
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posted:10/19/2009
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