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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 ≠ 0)
a cos2 u + b cos u + c = 0 ( a ≠ 0)
atg 2 u + btgu = c = 0 ( a ≠ 0)
a cot g 2 u + b cot gu + c = 0 ( a ≠ 0)
Caù c h giaû i:
Ñaët : t = sin u hay t = cos u vôù i t ≤ 1
π
t = tgu (ñieà u kieä n u ≠ + kπ )
2
t = cot gu (ñieà u kieä n u ≠ kπ )
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 = ± ≠− )
3 2 2
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 ( *)
1 + cos 6x 1 + cos 2x
Ta coù : (*) ⇔ .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
kπ
⇔ 2x = kπ ⇔ x = ( k ∈ Z)
2
1
Caù c h 2: (**) ⇔ ( cos 8x + cos 4x ) − 1 = 0
2
⇔ cos 8x + cos 4x − 2 = 0
⇔ 2 cos2 4x + cos 4x − 3 = 0
⎡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 :
(*)
1⎡ π⎞ ⎤ 3
( )
2 ⎛
⇔ sin2 x + cos2 x − 2 sin2 x cos2 x +
⎢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
2 2
( 2
)
⇔ − sin2 2x − 1 − 2 sin2 2x + sin 2x − = 0
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
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
sin2 x
Khi ñoù: (*) ⇔ 5 sin x − 2 = 3 (1 − sin x )
cos2 x
sin2 x
⇔ 5sin x − 2 = 3 (1 − sin x )
1 − sin2 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 )
⎣
π 5π
⇔x= + k2π ∨ x = + k2π ( k ∈ Z)
6 6
1 1
Baø i 60: Giaûi phöông trình: 2 sin 3x − = 2 cos 3x + ( *)
sin x cos x
Ñieà u kieä n : sin 2x ≠ 0
1 1
Luù c ñoù : (*) ⇔ 2 ( sin 3x − cos 3x ) = +
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 ⎦
⎡sin x + cos x = 0 ⎡ tgx = −1
⇔⎢ 2
⇔⎢
⎢sin 2x = 1 ∨ sin 2x = −1
( nhaän so vôùi ñieàu kieän )
⎣4 sin 2x − 2sin 2x − 2 = 0 ⎣ 2
π π π 7π
⇔x=− + 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
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
1 1 1
Ta coù : (*) ⇔ 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 =
⎣ 2 ⎢ x = π + k2π ∨ x = 5π + k2π
⎢
⎣ 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 )
⎣
π 2 π
⇔x= + kπ ∨ sin x = = sin
2 2 4
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⎠
π
(*) ⇔ 2 cos 2x.cos + 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 ⇔ ⎢
2
⎢sin x = 1
⎢
⎣ 2
π 5π
⇔ x = + k2π hay x = + k2π, k ∈
6 6
( )
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
(*) ⇔ 3
sin x 4
+2 2 = 2+3 2 ( )
sin2 x
vaø sin x ≠ 0
cos x
Ñaët t = ta ñöôï c phöông trình:
sin 2 x
( )
3t 2 − 2 + 3 2 t + 2 2 = 0
2
⇔t= 2∨t=
3
2 cos x 2
* Vôù i t = ta coù : 2
=
3 sin x 3
(
⇔ 3 cos x = 2 1 − cos2 x )
⇔ 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
⎢cos x = ( nhaän do cos x ≠ ±1)
⎣ 2
π
⇔ x = ± + k2π, k ∈
4
4 sin2 2x + 6 sin 2 x − 9 − 3 cos 2x
Baø i 66 : Giaûi phöông trình: = 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
1
⇔ cos 2x = −1 ∨ cos 2x = −
2
1
⇔ 2 cos2 x − 1 = −1 ∨ 2 cos2 x − 1 = −
2
⎡cos x = 0 ( loaïi do ñieàu kieän )
⇔⎢⎢cos x = ± 1 nhaän do cos x ≠ 0
⎢ ( )
⎣ 2
π 2π
⇔ x = ± + k2π ∨ x = ± + k2π ( k ∈ Z )
3 3
1 2
Baø i 67: Cho f ( x ) = sin x + sin 3x + sin 5x
3 5
Giaû i phöông trình: f ' ( x ) = 0
Ta coù : f '(x) = 0
⇔ cos x + cos 3x + 2 cos 5x = 0
⇔ ( 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 2x = ∨ 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
- 17
Baø i 68: Giaûi phöông trình: sin8 x + cos8 x = cos2 2x ( *)
16
Ta coù :
( )
2
sin 8 x + cos8 x = sin4 x + cos4 x − 2 sin 4 x cos4 x
2
1
( )
2
= ⎡ sin 2 x + cos2 x
⎢ − 2 sin 2 x cos2 x ⎤ − sin4 2x
⎥
⎣ ⎦ 8
2
⎛ 1 ⎞ 1
= ⎜ 1 − sin2 2x ⎟ − sin 4 2x
⎝ 2 ⎠ 8
1
= 1 − sin2 2x + sin4 2x
8
Do ñoù :
1
( *) ⇔ 16 ⎛ 1 − sin2 2x +
⎜
⎝ 8
⎞
(
sin4 2x ⎟ = 17 1 − sin2 2x
⎠
)
⇔ 2 sin4 2x + sin2 2x − 1 = 0
⎡sin2 2x = −1 ( loaïi )
1 1
⇔⎢ ⎢sin2 2x = 1 ⇔ (1 − cos 4x ) =
2 2
⎢
⎣ 2
π
⇔ cos 4x = 0 ⇔ x = ( 2k + 1) , ( k ∈ Z )
8
5x x
Baø i 69 : Giaûi phöông trình: sin = 5 cos3 x.sin ( *)
2 2
x
Nhaän xeù t thaáy : cos = 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
x
⇔ 3sin x − 4 sin3 x + 2 sin x cos x = 5 cos3 x.sin x vaø cos ≠0
2
⎧ x
⎪cos ≠ 0
⇔⎨ 2
⎪3 − 4 sin2 x + 2 cos x = 5 cos3 x ∨ sin x = 0
⎩
- ⎧ x
⎪ cos ≠ 0
⎪ 2
⇔ ⎨
⎪5 cos3 x − 4 cos2 x − 2 cos x + 1 = 0 ∨ sin x = 0
⎪
⎩ 2
⎧cos x ≠ −1
⎪
⇔ ⎨ x
( )
⎪( cos x − 1) 5 cos x + cos x − 1 = 0 ∨ sin 2 = 0
⎩
2
⎧cos x ≠ −1
⎪
⎪⎡
⎪ ⎢cos x = 1
⎪⎢
⇔ ⎨⎢ −1 + 21
⎪ ⎢cos x = 10
= cos α
⎪⎢
⎪⎢ −1 − 21
⎪ ⎣cos x =
⎢ 10
= cos β
⎩
⇔ 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
1
⇔ cos 2x = −1 ∨ cos 2x = ( 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
6x 8x
Baø i 71 : Giaûi phöông trình: 2 cos2 + 1 = 3 cos ( *)
5 5
- ⎛ 12x ⎞ ⎛ 2 4x ⎞
Ta coù : (*) ⇔ ⎜ 1 + cos ⎟ + 1 = 3 ⎜ 2 cos − 1⎟
⎝ 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
1 − 21 1 + 21
⇔ t = 1∨ t = ∨t = ( loïai )
4 4
Vaä y
4x 4x
• cos =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
⎛ π⎞
Baø i 72 : Giaûi phöông trình tg3 ⎜ x − ⎟ = tgx − 1 ( *)
⎝ 4⎠
π π
Ñaë t t = x − ⇔ x = + t
4 4
⎛π ⎞ 1 + tgt
(*) thaø n h : tg3 t = tg ⎜ + t ⎟ − 1 = − 1 vôùi cos t ≠ 0 ∧ tgt ≠ 1
⎝4 ⎠ 1 − tgt
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 )
π
⇔ t = kπ ∨ t = − + kπ, k ∈¢
4
Vaä y (*)
- π
⇔x= + kπ hay x = kπ, k ∈¢
4
sin 4 2x + cos4 2x
Baø i 73 : Giaûi phöông trình = cos4 4x (*)
⎛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
Baø i 74 :Giaû i phöông trình: 48 − − 2 (1 + cot g2x cot gx ) = 0 ( *)
cos x sin x
4
Ñieà u kieä n : sin 2x ≠ 0
Ta coù :
- cos 2x cos x
1 + cot g2x cot gx = 1 + .
sin 2x sin x
sin 2x sin x + cos 2x cos x
=
sin x sin 2x
cos x 1
= = ( do cos x ≠ 0 )
2 sin x cos x 2 sin 2 x
2
1 1
Luù c ñoù (*) ⇔ 48 − − 4 =0
cos x sin x
4
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
⇔ 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 : (*)
5
( ) (
⇔ sin8 x − 2 sin10 x + cos8 x − 2 cos10 x = ) 4
cos 2x
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
⇔ 4 cos 2x ( sin x − cos x ) = 5 cos 2x
8 8
- ⇔ 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 ⎠
⇔ cos 2x = 0 hay − 2 sin 2x = 1(Voâ nghieäm )
2
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
Luù c ñoù tg2x = , sin 2x = , cos 2x =
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
1 1 + t 2 + 1 ⎡1 − 1 − t ⎤ − 1 . 2t
2
−1 = ⎢ ⎥
t 1+t 2⎣ 1 + t2 ⎦ 2 1 + t2
1−t 1 − t 1 2t 2 t
⇔ = + . − ( do t ≠ −1)
t 1+t 2
2 1+t 2
1 + t2
2
1 − t t 2 − 2t + 1 (1 − t )
⇔ = =
t 1 + t2 1 + t2
⇔ ( 1 − t ) (1 + t 2 ) = ( 1 − t ) t
2
⎡1 − t = 0 ⎡ t = 1 ( nhaän do t ≠ −1)
⇔⎢ ⇔⎢ 2
⎣1 + t = (1 − t ) t ⎢2t − t + 1 = 0 ( voâ nghieäm )
2
⎣
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
2
cot gx − tgx + 4 sin 2x = ( *)
sin 2x
Ñieà u kieä n : sin 2x ≠ 0
2t
Ñaë t t = tgx thì : sin 2x = do sin 2x ≠ 0 neân t ≠ 0
1 + t2
1 8t 1 + t2 1
(*) thaø n h : − t + = = +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 )
⎛ π⎞
Vaäy (*) ⇔ 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 ⎠
( t + 1) = 1 + t
2
⇔ (1 − t )
1 + t2
⎡ t = −1
⎡ t = −1
⇔ ⎢ (1 − t )(1 + t ) ⇔ ⎢
⎢ =1 ⎣1 − t = 1 + t
2 2
⎢
⎣ 1+t 2
⇔ t = −1 ∨ t = 0
- ⎡ tgx = −1 π
Do ñoù (*) ⇔ ⎢ ⇔ x = − + kπ hay x = kπ, k ∈
⎣ tgx = 0 4
Baø i 80 : Cho phöông trình cos 2x − ( 2m + 1) cos x + m + 1 = 0 ( * )
3
a/ Giaû i phöông trình khi m =
2
⎛ π 3π ⎞
b/ Tìm m ñeå (*) coù nghieä m treâ n ⎜ , ⎟
⎝2 2 ⎠
Ta coù (*) 2 cos x − ( 2m + 1) cos x + m = 0
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
⎡ 2π ⎤
b/ Tìm m sao cho (*) coù ñuù n g hai nghieä m treâ n ⎢0, ⎥
⎣ 3⎦
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
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
2
Baø i 82 : Cho phöông trình (1 − a ) tg 2 x − + 1 + 3a = 0 (1)
cos x
1
a/ Giaû i (1) khi a =
2
⎛ 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⎞
a/ Khi a = 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)
Ta coù : (1) ⇔ ⎢ 2
⎢
⎢2a cos x = 1 − a ( 2 )
⎣
⎧
⎪a ≠ 0
⎪
⎧1 ⎫ ⎪ 1−a
Yeâ u caà u baø i toaù n ⇔ (2) coù nghieä m treâ n ( 0,1) \ ⎨ ⎬ ⇔ ⎨0 < 0 ⎧1
⎪ 2a
⎪
⎪ 1 ⎪3 < a < 1
⎪
⇔⎨ ⇔ ⎨a < 0 ∨ a > ⇔ ⎨
⎪ 1 − 3a ⎪ 3 ⎪a ≠ 1
- ⎧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]
17
⇔2 ≤ m <
8
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
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
x − 3x + 2 < 0
4 2
- (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
⇔ 4x = kπ ∨ 4x = + 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
⇔ − 2 < x < −1 ∨ 1 < x < 2 ( *)
π ⎛ π⎞
x=− thì sin 4x = sin ⎜ − ⎟ = −1
8 ⎝ 2⎠
π
x = − laø nghieäm cuûa (1) ⇒ 1 + 1 + m = 0
8
⇒ m = −2
Luù c ñoù (1) thaø nh : sin2 4x − sin 4x − 2 = 0
⎧t = sin 4x ( vôùi t ≤ 1)
⎪
⇔⎨
⎪t − t − 2 = 0
2
⎩
⎧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 ) ( 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
⇔ cos x = 0 ∨ cos x =
2
( )
Ta coù : (2) ⇔ 4 cos x − 4 cos x − 3 cos x = a cos x + ( 4 − a ) 2 cos2 x
2 3
⇔ 4 cos3 x + ( 4 − 2a ) cos2 x ( a − 3) cos x = 0
⎡cos x = 0
⇔⎢
⎢4 cos x + 2 ( 2 − a ) cos x + a − 3 = 0
2
⎣
⎛ 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
⇔ ⎢a − 3 = 1 ⇔ ⎢a = 4
⎢ 2 2 ⎢
⎢a − 3 a−3 ⎣a < 1 ∨ a > 5
⎢
⎢ < −1 ∨ >1
⎣ 2
⎢ 2
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 2 cos 2x − 1 = 1 + 4 cos 2x − 3 cos 2x + a (1 − cos 2x )
2 3
⎧t = cos 2x
⎪ ( t ≤ 1)
⇔⎨
⎩ ( 2
)
⎪2 2t − 1 = 1 + 4t − 3t + a (1 − t )
3
⎧t = cos 2x
⎪ ( t ≤ 1)
⇔⎨
⎪−4t + 4t + 3t − 3 = a (1 − t )
3 2
⎩
⎧1 = cos 2x
⎪ ( t ≤ 1)
⇔⎨
( )
⎪( t − 1) −4t + 3 = a (1 − t ) ( * *)
⎩
2
a/ Khi a = 1 thì (*) thaø nh :
nguon tai.lieu . vn
