Chương 2: Phương trình lượng giác cơ bản

 Chöông 2: PHÖÔNG TRÌNH LÖÔÏNG GIAÙC CÔ BAÛN 
= + π⎡= ⇔ ⎢ = π − + π⎣
u v k2
sin u sin v
u v k2
cosu cosv u v k2= ⇔ = ± + π 
π⎧ ≠ + π⎪= ⇔ ⎨⎪ = + π⎩
u k
tgu tgv 2
u v k '
 ( )k,k ' Z∈ 
u k
cot gu cot gv
u v k '
≠ π⎧= ⇔ ⎨ = + π⎩ 
Ñaëc bieät : si nu 0 u k= ⇔ = π π= ⇔ = + πco su 0 u k
2
(sinu 1 u k2 k Z
2
π= ⇔ = + π ∈ ) cosu 1 u k2= ⇔ = π ( ) k Z∈
sinu 1 u k2
2
π= − ⇔ = − + π cosu 1 u k2= − ⇔ = π + π 
Chuù yù : sinu 0 cosu 1≠ ⇔ ≠ ±
cosu 0 sinu 1≠ ⇔ ≠ ± 
Baøi 28 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2002) 
[ ]x 0,14∈ nghieäm ñuùng phöông trình Tìm 
( )cos3x 4 cos2x 3cos x 4 0 *− + − = 
Ta coù (*) : ⇔ ( ) ( )3 24 cos x 3cos x 4 2cos x 1 3cos x 4 0− − − + − = 
⇔ 3 24 cos x 8cos x 0− = ⇔ ( )24 cos x cos x 2 0− = 
⇔ ( )= =cos x 0hay cos x 2 loaïi vì cos x 1≤ 
⇔ ( )x k k
2
π= + π ∈ Z 
Ta coù : [ ]x 0,14 0 k 1
2
4π∈ ⇔ ≤ + π ≤ 
⇔ k 14
2 2
π π− ≤ π ≤ − ⇔ 1 14 10,5 k 3,9
2 2
− = − ≤ ≤ − ≈π 
 Maø k neân Z∈ { }k . Do ñoù : 0,1,2,3∈ 3 5 7x , , ,
2 2 2 2
π π π π⎧ ⎫∈ ⎨ ⎬⎩ ⎭ 
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 *− + = − 
Ta coù (*) ⇔ ( ) ( ) ( )− + =2cos x 1 2sin x cos x sin x 2cos x 1− 
⇔ ( ) ( )2cos x 1 2sin x cos x sin x 0− + −⎡ ⎤⎣ ⎦ =
)
⇔ ( ) (2cos x 1 sin x cos x 0− + = 
⇔ 1cos x sin x cos x
2
= ∨ = − 
⇔ cos x cos tgx 1 tg
3 4
π π⎛ ⎞= ∨ = − = −⎜ ⎟⎝ ⎠ 
⇔ ( )π π= ± + π ∨ = − + π ∈x k2 x k , k
3 4
Z 
Baøi 30 : Giaûi phöông trình + + + =cosx cos2x cos3x cos4x 0(*) 
Ta coù (*) ⇔ ( ) ( )cos x cos4x cos2x cos3x 0+ + + = 
⇔ 5x 3x 5x x2cos .cos 2cos .cos 0
2 2 2 2
+ = 
⇔ 5x 3x x2cos cos cos 0
2 2 2
⎛ ⎞+ =⎜ ⎟⎝ ⎠ 
⇔ 5x x4cos cos x cos 0
2 2
= 
⇔ 5x xcos 0 cos x 0 cos 0
2 2
= ∨ = ∨ = 
⇔ π π π= + π ∨ = + π ∨ = + π5x xk x k k
2 2 2 2 2
⇔ ( )π π π= + ∨ = + π ∨ = π + π ∈2kx x k x 2 ,
5 5 2
k Z 
Baøi 31: Giaûi phöông trình ( )2 2 2 2sin x sin 3x cos 2x cos 4x *+ = + 
Ta coù (*) ⇔ ( ) ( ) ( ) ( )1 1 1 11 cos2x 1 cos6x 1 cos4x 1 cos8x
2 2 2 2
− + − = + + + 
⇔ ( )cos2x cos6x cos4x cos8x− + = +
⇔ 2cos4xcos2x 2cos6xcos2x− =
⇔ ( )2cos2x cos6x cos4x 0+ = 
⇔ 4cos2xcos5xcos x 0= 
⇔ cos2x 0 cos5x 0 cosx 0= ∨ = ∨ = 
⇔ π π π= + π ∨ + π ∨ = + π ∈ 2x k 5x k x k ,k
2 2 2
⇔ π π π π π= + ∨ = + ∨ = + πkk kx x x
4 2 10 5 2
 ∈ , k 
Baøi 32 : Cho phöông trình 
( )π⎛ ⎞− = − −⎜ ⎟⎝ ⎠
2 2 x 7sin x.cos4x sin 2x 4sin *
4 2 2
Tìm caùc nghieäm cuûa phöông trình thoûa: − <x 1 3 
Ta coù : (*)⇔ ( )1 7sin x.cos4x 1 cos4x 2 1 cos x
2 2
⎡ π ⎤⎛ ⎞
2
− − = − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ − 
⇔ − + = − −1 1 3sin x cos4x cos4x 2sin x
2 2 2
⇔ 1sin x cos4x cos4x 1 2sin x 0
2
+ + + = 
⇔ ⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1cos4x sin x 2 sin x 0
2 2
⇔ ( ) 1cos4x 2 sin x 0
2
⎛ ⎞+ + =⎜ ⎟⎝ ⎠ 
⇔ 
( )cos4x 2 loaïi
1sin x sin
2 6
= −⎡⎢ π⎛⎢ = − = −⎜ ⎟⎢ ⎝ ⎠⎣
⎞ ⇔ 
π⎡ = − + π⎢⎢ π⎢ = + π⎢⎣
x k
6
7x 2
6
2
h
Ta coù : − <x 1 3 ⇔ ⇔ 3 x 1 3− < − < 2 x 4− < < 
Vaäy : 2 k2
6
π− < − + π < 4 
⇔ 2 2k 4
6 6
π π− < π < + ⇔ 1 1 2 1k
12 12
− < < +π π 
Do k neân k = 0. Vaäy Z∈ x
6
π= − 
π− < + π <72 h2
6
4 
 ⇔ π π− − < π < − ⇔ − − < < −π π
7 7 1 7 22 h2 4 h
6 6 12
7
12
⇒ h = 0 ⇒ π= 7x
6
.Toùm laïi −π π= = 7x hay x
6 6
Caùch khaùc : −π= − ⇔ = − + π ∈ k1sin x x ( 1) k , k
2 6
Vaäy : 
−π − −− < − + π < ⇔ < − + <π π
k k2 12 ( 1) k 4 ( 1) k
6 6
4 
⇔ k=0 vaø k = 1. Töông öùng vôùi −π π= = 7x hay x
6 6
Baøi 33 : Giaûi phöông trình 
( )3 3 3sin x cos3x cos x sin 3x sin 4x *+ = 
Ta coù : (*)⇔ ( ) ( )3 3 3 3 3sin x 4 cos x 3cos x cos x 3sin x 4sin x sin 4x− + − = 
⇔ 3 3 3 3 3 3 34sin xcos x 3sin xcosx 3sin xcos x 4sin xcos x sin 4x− + − =
⇔ ( )2 2 33sin x cos x cos x sin x sin 4x− =
⇔ 33 sin2x cos2x sin 4x
2
= 
⇔ 33 sin4x sin 4x
4
= 
⇔ 33sin4x 4sin 4x 0− = 
⇔ sin12x = 0 
⇔ ⇔ 12x k= π ( )kx k
12
Zπ= ∈ 
Baøi 34 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B, naêm 2002) 
Giaûi phöông trình : 
( )2 2 2 2sin 3x cos 4x sin 5x cos 6a *− = − 
Ta coù : (*)⇔ 
( ) ( ) ( ) ( )1 1 1 11 cos6x 1 cos8x 1 cos10x 1 cos12x
2 2 2 2
− − + = − − + 
⇔ cos6x cos8x cos10x cos12x+ = +
⇔ 2cos7xcosx 2cos11xcosx=
⇔ ( )2cos x cos7x cos11x 0− = 
⇔ cosx 0 cos7x cos11x= ∨ =
⇔ π= + π ∨ = ± + πx k 7x 11x k
2
2 
⇔ π π π= + π ∨ = − ∨ = ∈ k kx k x x , k
2 2 9
Baøi 35 : Giaûi phöông trình 
( ) ( )sin x sin 3x sin 2x cos x cos3x cos2x+ + = + + 
⇔ 2sin2xcosx sin2x 2cos2xcosx cos2x+ = +
⇔ ( ) ( )+ = +sin 2x 2 cos x 1 cos 2x 2 cos x 1 
⇔ ( ) ( )2cos x 1 sin 2x cos2x 0+ − = 
⇔ 1 2cos x cos sin2x cos2x
2 3
π= − = ∨ = 
⇔ 2x k2 tg2x 1
3 4
tgπ π= ± + π ∨ = = 
⇔ ( )π π π= ± + π ∨ = + ∈2x k2 x k , k
3 8 2
Z 
Baøi 36: Giaûi phöông trình 
( )+ + = +2 3cos10x 2 cos 4x 6 cos 3x.cos x cos x 8 cos x. cos 3x * 
Ta coù : (*)⇔ ( ) ( )3cos10x 1 cos8x cos x 2cos x 4 cos 3x 3cos3x+ + = + − 
⇔ ( )cos10x cos8x 1 cos x 2cos x.cos9x+ + = +
⇔ 2cos9xcosx 1 cos x 2cosx.cos9x+ = +
⇔ cosx 1=
⇔ ( )x k2 k Z= π ∈ 
Baøi 37 : Giaûi phöông trình 
( )3 3 24 sin x 3cos x 3sin x sin x cos x 0 *+ − − = 
Ta coù : (*) ⇔ ( ) ( )2 2sin x 4sin x 3 cos x sin x 3cos x 02− − − = 
⇔ ( ) ( )⎡ ⎤− − − − =⎣ ⎦2 2sin x 4 sin x 3 cos x sin x 3 1 sin x 02
=
=
⇔ ( ) ( )24 sin x 3 sin x cos x 0− −
⇔ ( ) ( )2 1 cos2x 3 sin x cos x 0− − −⎡ ⎤⎣ ⎦
⇔ 
1 2cos2x cos
2 3
sin x cos x
π⎡ = − =⎢⎢ =⎣
⇔ 
22x k2
3
tgx 1
π⎡ = ± + π⎢⎢ =⎣
 ⇔ 
x k
3
x k
4
π⎡ = ± + π⎢⎢ π⎢ = + π⎢⎣
 ( )k Z∈ 
Baøi 38 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B naêm 2005) 
Giaûi phöông trình : 
( )sin x cos x 1 sin 2x cos2x 0 *+ + + + = 
Ta coù : (*) ⇔ 2sin x cosx 2sin xcosx 2cos x 0+ + + =
⇔ ( )sin x cos x 2cos x sin x cos x 0+ + + = 
⇔ ( ) (sin x cos x 1 2cos x 0+ + ) =
⇔ 
sin x cos x
1 2cos2x cos
2 3
= −⎡⎢ π⎢ = − =⎣
⇔ 
tgx 1
2x k
3
= −⎡⎢ π⎢ = ± + π⎣ 2
⇔ 
x k
4
2x k2
3
π⎡ = − + π⎢⎢ π⎢ = ± + π⎢⎣
( )k Z∈ 
Baøi 39 : Giaûi phöông trình 
( ) ( ) ( )22sin x 1 3cos4x 2sin x 4 4 cos x 3 *+ + − + = 
Ta coù : (*) ⇔ ( ) ( ) ( )22sin x 1 3cos4x 2sin x 4 4 1 sin x 3 0+ + − + − − = 
⇔ ( ) ( ) ( ) ( )2sin x 1 3cos4x 2sin x 4 1 2sin x 1 2sin x 0+ + − + + − = 
⇔ ( ) ( )2sin x 1 3cos4x 2sin x 4 1 2sin x 0+ + − + −⎡ ⎤⎣ ⎦ =
=
⇔ ( ) ( )3 cos4x 1 2sin x 1 0− +
⇔ 1cos4x 1 sin x sin
2 6
π⎛ ⎞= ∨ = − = −⎜ ⎟⎝ ⎠ 
⇔ π π= π ∨ = − + π ∨ = +74x k2 x k2 x k2
6 6
π 
⇔ ( )π π π= ∨ = − + π ∨ = + π ∈k 7x x k2 x k2 , k
2 6 6
Z 
Baøi 40: Giaûi phöông trình ( ) ( )+ = +6 6 8 8sin x cos x 2 sin x cos x * 
Ta coù : (*) ⇔ 6 8 6 8sin x 2sin x cos x 2cos x 0− + − = 
⇔ ( ) ( )6 2 6 2sin x 1 2sin x cos x 2cos x 1 0− − − = 
⇔ − =6 6sin x cos2x cos x.cos 2x 0
⇔ ( )6 6cos2x sin x cos x 0− =
⇔ 6 6cos2x 0 sin x cos x= ∨ =
⇔ 6cos2x 0 tg x 1= ∨ = 
⇔ ( )2x 2k 1 tgx 1
2
π= + ∨ = ± 
⇔ ( )x 2k 1 x k
4 4
π π= + ∨ = ± + π 
⇔ kx
4 2
π π= + ,k ∈ 
Baøi 41 : Giaûi phöông trình 
( )1cos x.cos2x.cos4x.cos8x *
16
= 
Ta thaáy x k= π khoâng laø nghieäm cuûa (*) vì luùc ñoù 
cosx 1,cos2x cos4x cos8x 1= ± = = = 
(*) thaønh : 11
16
± = voâ nghieäm 
Nhaân 2 veá cuûa (*) cho 16sin x 0≠ ta ñöôïc 
(*)⇔ vaø ( )16sin x cos x cos2x.cos4x.cos8x sin x= sin x 0≠
⇔ vaø ( )8sin 2x cos2x cos4x.cos8x sin x= sin x 0≠ 
⇔ vaø si( )4sin 4x cos4x cos8x sin x= n x 0≠ 
⇔ vaø 2sin8xcos8x sin x= sin x 0≠ 
⇔ sin16x sin x= vaø sin x 0≠ 
⇔ ( )π π π= ∨ = + ∈k2 kx x , k
15 17 17
Z 
Do : khoâng laø nghieäm neân = πx h ≠k 15m vaø ( )+ ≠ ∈2k 1 17n n, m Z
Baøi 42: Giaûi phöông trình ( )38cos x cos 3x *
3
π+ =⎛ ⎞⎜ ⎟⎝ ⎠ 
Ñaët t x x t
3 3
π π= + ⇔ = − 
Thì ( ) ( )cos 3x cos 3t cos 3t cos 3t= − π = π − = − 
Vaäy (*) thaønh = −38cos t cos3t
⇔ 3 38 cos t 4 cos t 3cos t= − +
⇔ 312cos t 3cos t 0− = 
⇔ ( )23 cos t 4 cos t 1 0− =
⇔ ( )3cos t 2 1 cos 2t 1 0+ −⎡ ⎤⎣ ⎦ =
⇔ ( )cos t 2 cos 2t 1 0+ =
⇔ 1 2cos t 0 cos 2t cos
2 3
π= ∨ = − = 
⇔ ( ) π π= + ∨ = ± +2t 2k 1 2t k2
2 3
π 
⇔ π π= + π ∨ = ± + πt k t
2 3
k 
Maø x t
3
π= − 
Vaäy (*)⇔ ( )π π= + π ∨ = π ∨ = + π ∈2x k2 x k x k , vôùik
6 3
Z 
Ghi chuù : 
Khi giaûi caùc phöông trình löôïng giaùc coù chöùa tgu, cotgu, coù aån ôû maãu, hay 
chöùa caên baäc chaün... ta phaûi ñaët ñieàu kieän ñeå phöông trình xaùc ñònh. Ta seõ 
duøng caùc caùch sau ñaây ñeå kieåm tra ñieàu kieän xem coù nhaän nghieäm hay 
khoâng. 
 + Thay caùc giaù trò x tìm ñöôïc vaøo ñieàu kieän thöû laïi xem coù thoûa 
Hoaëc + Bieåu dieãn caùc ngoïn cung ñieàu kieän vaø caùc ngoïn cung tìm ñöôïc treân cuøng 
moät ñöôøng troøn löôïng giaùc. Ta seõ loaïi boû ngoïn cung cuûa nghieäm khi coù 
truøng vôùi ngoïn cung cuûa ñieàu kieän. 
Hoaëc + So vôi caùc ñieàu kieän trong quaù trình giaûi phöông trình. 
Baøi 43 : Giaûi phöông trình ( )2tg x tgx.tg3x 2 *− = 
Ñieàu kieän 3
cos x 0
cos 3x 4 cos x 3cos x 0
≠⎧⎨ = − ≠⎩
π π⇔ ≠ ⇔ ≠ + hcos3x 0 x
6 3
Luùc ñoù ta coù (*) ⇔ ( )tgx tgx tg3x 2− = 
⇔ sin x sin x sin 3x 2
cos x cos x cos 3x
⎛ ⎞− =⎜ ⎟⎝ ⎠ 
⇔ ( ) 2sin x sin x cos 3x cos x sin 3x 2 cos x cos 3x− =
⇔ ( ) 2sin x sin 2x 2 cos x.cos 3x− = 
⇔ 2 22sin x cos x 2cos x cos3x− =
⇔ (do cos2sin x cos x cos3x− = x 0≠ ) 
⇔ ( ) ( )1 11 cos 2x cos 4x cos 2x
2 2
− − = + 
⇔ cos4x 1 4x k2= − ⇔ = π + π 
⇔ ( )kx k
4 2
π π= + ∈ Z 
so vôùi ñieàu kieän 
Caùch 1 : Khi 
kx
4 2
π= + π thì ( )3 3k 2cos 3x cos 0 nhaän
4 2 2
π π⎛ ⎞= + = ± ≠⎜ ⎟⎝ ⎠ 
Caùch 2 : Bieåu dieãn caùc ngoïn cung ñieàu kieän vaø ngoïn cung nghieäm ta thaáy 
khoâng coù ngoïn cung naøo truøng nhau. Do ñoù : 
(*) ⇔ kx
4 2
π π= + 
Löu yù caùch 2 raát maát thôøi gian 
Caùch 3 : 
Neáu 
π π π= + = +3 3k3x h
4 2 2
π
h 6k
Thì + = +3 6k 2 4h
⇔1 4 = −
⇔ = −1 2h 3k
2
 (voâ lyù vì ∈k,h Z ) 
Baøi 44: Giaûi phöông trình 
( )2 2 2 11tg x cot g x cot g 2x *
3
+ + = 
Ñieàu kieän 
cos x 0
sin x 0 sin 2x 0
sin 2x 0
≠⎧⎪ ≠ ⇔ ≠⎨⎪ ≠⎩
Do ñoù : 
(*)⇔ 2 2 21 1 11 1 1cos x sin x sin 2x 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
11 
⇔ 2 2 2 21 1 1cos x sin x 4 sin x cos x 3+ + =
20
⇔
2 2
2 2
4 sin x 4 cos x 1 20
4 sin x cos x 3
+ + = 
⇔ 25 2sin 2x 3=
0
⇔ 2 3sin 2x
4
= (nhaän do sin2x 0≠ ) 
⇔ ( )1 31 cos4x
2 4
− = 
⇔ 1 2cos4x cos
2 3
π= − = 
⇔ 24x k2
3
π= ± + π 
⇔ ( )kx k
6 2
π π= ± + ∈ Z 
Chuù yù : Coù theå deã daøng chöùng minh : 
2tgx cot gx
sin 2x
+ = 
Vaäy (*)⇔( )2 21 1tgx cot gx 2 1sin x 3
⎛ ⎞+ − + − =⎜ ⎟⎝ ⎠
1 
⇔ 25 2sin 2x 3=
0
Baøi 45 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2003) 
Giaûi phöông trình 
( )2 2 2x xsin tg x cos 0 *
2 4 2
π⎛ ⎞− − =⎜ ⎟⎝ ⎠ 
Ñieàu kieän : cos x 0 sin x 1≠ ⇔ ≠ ±
luùc ñoù : 
(*) ⇔ [ ]221 sin x 11 cos x 1 cos x 02 2 cos x 2
⎡ π ⎤⎛ ⎞− − − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ = 
 ⇔ ( ) ( ) ( )221 sin x 1 cos x 1 cos x 01 sin x
− − − + =− 
⇔ ( )21 cos x 1 cos x 0
1 sin x
− − + =+ 
⇔ ( ) 1 cos x1 cos x 1 0
1 sin x
−⎡ ⎤+ −⎢ ⎥+⎣ ⎦ =
=
⇔ ( ) ( )1 cos x cos x sin x 0+ − −
⇔ ( )cos x 1 nhaändocos x 0
tgx 1
= − ≠⎡⎢ = −⎣
⇔ 
= π + π⎡⎢ π⎢ = − + π⎣
x k2
x k
4
Baøi 46 : Giaûi phöông trình 
( ) ( )2sin 2x cot gx tg2x 4 cos x *+ = 
Ñieàu kieän : ⇔ sin x 0
cos2x 0
≠⎧⎨ ≠⎩ 2
sin x 0
2cos x 1 0
≠⎧⎨ − ≠⎩
 ⇔ 
cos x 1
2cos x
2
≠ ±⎧⎪⎨ ≠ ±⎪⎩
Ta coù : cos x sin2xcot gx tg2x
sin x cos2x
+ = + 
cos2x cos x sin2xsin x
sin x cos2x
+= 
cos x
sin x cos2x
= 
Luùc ñoù : (*) ⎛ ⎞⇔ =⎜ ⎟⎝ ⎠
2cos x2sin x cos x 4 cos x
sin x cos 2x
⇔ 
2
22cos x 4 cos x
cos2x
= ( )Dosin x 0≠ 
⇔ 
cos x 0
1 2
cos2x
=⎡⎢⎢ =⎣
 ⇔ 
( )
⎡ ⎛ ⎞= ≠ ≠ ±⎢ ⎜ ⎟⎜ ⎟⎢ ⎝ ⎠⎢ π⎢ = = ≠⎣
2cos x 0 Nhaän do cos x vaø 1
2
1cos 2x cos , nhaän do sin x 0
2 3
⇔ 
π⎡ = + π⎢⎢ π⎢ = ± + π⎢⎣
x k
2
x k
6
 ( ) ∈k Z
Baøi 47 : Giaûi phöông trình: 
( )2 2cot g x tg x 16 1 cos4x
cos2x
− = + 
Ta coù : 
2 2
2 2
2 2
cos x sin xcot g x tg x
sin x cos x
− = − 
4 4
2 2 2
cos x sin x 4cos2x
sin x cos x sin 2x
−= = 
Ñieàu kieän : ⇔ sisin2x 0
cos2x 0
≠⎧⎨ ≠⎩ n4x 0≠ 
Luùc ñoù (*) ( )24 16 1 cos4xsin 2x⇔ = + 
( )
( ) ( )
( )
( )
( )
⇔ = +
⇔ = + −
⇔ = − =
⇔ = ≠
⇔ − =
π π⇔ = ⇔ = + ∈ 
2
2 2
2
1 4 1 cos 4x sin 2x
1 2 1 cos 4x 1 cos 4x
1 2 1 cos 4x 2sin 4x
1sin 4x nhaän do sin 4x 0
2
1 11 cos 8x
2 2
kcos 8x 0 x , k
16 8
Baøi 48: Giaûi phöông trình: ( )4 4 7sin x cos x cot g x cot g x *
8 3 6
π π⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 
Ñieàu kieän 
sin x 0 sin x 0
3 3 2sin 2x 0
3
sin x 0 cos x 0
6 3
⎧ ⎧π π⎛ ⎞ ⎛ ⎞+ ≠ + ≠⎜ ⎟ ⎜ ⎟⎪ ⎪ π⎪ ⎝ ⎠ ⎪ ⎝ ⎠ ⎛ ⎞⇔ ⇔ +⎨ ⎨ ⎜ ⎟π π ⎝ ⎠⎛ ⎞ ⎛ ⎞⎪ ⎪− ≠ + ≠⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎩
≠ 
1 3sin2x cos2x 0
2 2
tg2x 3
⇔ − + ≠
⇔ ≠
Ta coù: ( )24 4 2 2 2 2 21sin x cos x sin x cos x 2sin x.cos x 1 sin 2x2+ = + − = − 
Vaø: cot g x .cot g x cot g x .tg x 1
3 6 3 3
π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 
Luùc ñoù: (*) 21 71 sin 2x
2 8
⇔ − = 
 ( )1 11 cos4x
4 8
⇔ − − = − 
⇔ =
π π⇔ = ± + π ⇔ = ± +
1cos 4x
2
k4x k2 x
3 1
π
2 2
 (nhaän do 3tg2x 3
3
= ± ≠ ) 
Baøi 49: Giaûi phöông trình ( )12tgx cot g2x 2sin2x *
sin2x
+ = + 
Ñieàu kieän: 
cos2x 0
sin2x 0 cos2x 1
sin2x 0
≠⎧ ⇔ ≠ ⇔ ≠ ±⎨ ≠⎩ 
Luùc ñoù: (*) 
2sin x cos2x 12sin2x
cos x sin2x sin2x
⇔ + = + 
( )
( )
( )
( )
( )
( )
⇔ + = +
⇔ + − = +
⇔ − =
⎡ ⎤⇔ − + =⎣ ⎦
⎡ = ≠ ⇒⎢⇔ π⎢
≠
= − = ≠ ±⎢⎣
π⇔ = ± + π ∈
π⇔ = ± + π ∈ 
2 2
2 2 2 2
2 2
2
4 sin x cos 2x 2sin 2x 1
4 sin x 1 2sin x 8sin x cos x 1
2sin x 1 4 cos x 0
2sin x 1 2 1 cos 2x 0
sin x 0 loaïi do sin 2x 0 sin x 0
1 2cos 2x cos nhaän do cos 2x 1
2 3
22x k2 k Z
3
x k , k
3
Baøi 51: Giaûi phöông trình: 
( ) ( )3 sin x tgx 2 1 cos x 0 *
tgx sin x
+ − + =− ( ) 
Ñieàu kieän : ⇔ tgx sin x 0− ≠ sin x sin x 0
cos x
− ≠ 
⇔ ( )sin x 1 cos x 0
cos x
− ≠ ⇔ 
sin x 0
cos x 0 sin2x 0
cos x 1
≠⎧⎪ ≠ ⇔ ≠⎨⎪ ≠⎩
Luùc 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 2cosx 0+ =
⇔ 1cos x
2
= − (nhaän so vôùi ñieàu kieän) 
⇔ π= ± + π ∈ 2x k2 , k
3
Baøi 52 : Giaûi phöông trình 
( ) ( )
( ) ( ) ( )
2 2
2 21 cos x 1 cos x 1tg xsin x 1 sin x tg x *
4 1 sin x 2
− + + − = + +− 
Ñieàu kieän : 
cos x 0
sin x 1
≠⎧⎨ ≠⎩ ⇔ cos x 0≠ 
Luùc ñoù (*)⇔ ( )( ) ( )
2 3 2
2 2
2 1 cos x sin x 1 sin x1 sin x
4 1 sin x 1 sin x 2 1 sin x
+ − = + +− − − 
⇔ ( ) ( ) ( ) ( )2 3 21 cos x 1 sin x 2sin x 1 sin x 1 sin x 2sin x+ + − = + − + 2 
⇔ ( ) ( ) ( ) ( )2 2 21 sinx 1 cos x 1 sin x cos x 2sin x 1 sin x+ + = + + + 
⇔ 2 2
1 sin x 0
1 cos x cos x 2sin x
+ =⎡⎢ + = +⎣ 2
⇔ ⇔ cos2x = 0 = − ≠⎡⎢ = −⎣
sin x 1 ( loaïi do cos x 0 )
1 1 cos 2x
⇔ 2x k
2
π= + π 
⇔ x k
4 2
π= + π (nhaän do cosx ≠ 0) 
Baøi 53 : Giaûi phöông trình 
( )cos3x.tg5x sin7x *= 
Ñieàu kieän cos5x 0≠
Luùc ñoù : (*) ⇔ sin5xcos3x. sin7x
cos5x
= 
sin5x.cos3x sin7x.cos5x= ⇔ 
[ ] [ ]1 1sin8x sin2x sin12x sin2x
2 2
+ = + ⇔ 
sin8x sin12x= ⇔ 
 12x 8x k2 12x 8x k2= + π ∨ = π − + π⇔
π π= ∨ = +k kx x⇔ π
10
So laïi vôùi ñieàu kieän 
2 20
k 5kx thì cos5x cos cos
2 2
π π= = = k
2
π
 (loaïi neáu k leû) 
π π π π⎛ ⎞= + = + ≠⎜ ⎟⎝ ⎠20 10
k kx thì cos5x cos 0 nhaän
4 2
π π= π ∨ = + kx h x
20 10
Do ñoù : (*)⇔ , vôùi k, h ∈ 
Baøi 54 : Giaûi phöông trình 
( )4 4sin x cos x 1 tgx cot g2x *
sin2x 2
+ = + ( )
Ñieàu kieän :
Ta coù :
 sin2x 0≠ 
 ( )24 4 2 2 2sin x cos x sin x cos x 2sin x cos x+ = + − 2
211 sin 2
2
= − x
sin x cos2xtgx cot g2x
cos x sin2x
+ = + 
sin2xsin x cosx cos2x
cos xsin2x
+= 
( )cos 2x x 1
cos xsin2x sin2x
−= = 
( )
−
⇔ =
⇔ − =
⇔ = ≠
⇔ =
π⇔ = + π ∈
π π⇔ = + ∈


2
2
2
2
11 sin 2x 12Do ñoù : (*)
sin 2x 2sin 2x
1 11 sin 2x
2 2
sin 2x 1 nhaän do sin 2x 0
cos 2x 0
2x k , k 
2
kx , k 
4 2
Baøi 55 : Giaûi phöông trình 
( )2tg x x2 2 2.cot g 2x.cot g3x tg x cot g 2 cot g3x *= − + 
cos x 0 sin2x 0 sin3x 0≠ ∧ ≠ ∧ ≠ Ñieàu kieän : 
sin2x 0 sin3x 0⇔ ≠ ∧ ≠
( )⇔ − = −
⎡ − + ⎤ − +⎛ ⎞ ⎛ ⎞⇔ − =⎜ ⎟ ⎜ ⎟⎢ ⎥+ − + −⎝ ⎠ ⎝ ⎠⎣ ⎦
2 2 2 2Luùc ñoù (*) cotg3x tg x cot g 2x 1 tg x cot g 2x
1 cos 2x 1 cos 4x 1 cos 2x 1 cos 4xcot g3x 1
1 cos 2x 1 cos 4x 1 cos 2x 1 cos 4x
 −
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
[ ] ( )
[ ]
( )
⎡ ⎤⇔ − + − + −⎣ ⎦
= − − − + +
⇔ − = − +
⇔ − = −
⇔ = ≠
⇔ = ∨ =
π⇔ = + π ∨ =
cot g3x 1 cos2x 1 cos4x 1 cos2x 1 cos4x
1 cos2x 1 cos4x 1 cos4x 1 cos2x
cot g3x 2cos4x 2cos2x 2 cos4x cos2x
cos3x 4sin3xsin x 4cos3x cos x
sin3x
cos3xsin x cos3x cos x do sin3x 0
cos3x 0 sin x cos x
3x k tgx 1
2
( )π π π⇔ = + ∨x x
6 3
= + π ∈k l k, l Z
4
 ñieàu kieän: sinSo vôùi ≠2x.sin3x 0 
* Khi π π= + kx
6 3
 thì π π π⎛ ⎞ ⎛ ⎞+ + π ≠⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2ksin .sin k 0
3 3 2
+⎛ ⎞⇔ π ≠⎟⎠
1 2ksin 0 
Luoân ñuùng 
⎜⎝ 3
( )∀ + ≠k thoûa 2k 1 3m m Z ∈
* Khi π= + πx l thì π π⎛ ⎞ ⎛ ⎞+ π + π = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
3 2sin 2l sin 3l 0
2 4 2
4
≠
luoân ñuùng 
Do ñoù: (*) 
π π⎡ = + ∈ ∧ ≠ − ∈⎢⇔ ⎢ π⎢ = + π ∈⎢⎣


kx , k Z 2k 3m 1( m
6 3
x l , l
4
)
Caùch khaùc: ( )⇔ − = −
−−⇔ = =− −
+ −⇔ = − +
⇔ = ⇔ = ∨ =
2 2 2 2
2 22 2
2 2 2 2
 (*) cotg3x tg x cot g 2x 1 tg x cot g 2x
tg 2x.tg x 1tg x cot g 2xcot g3x
tg x cot g 2x 1 tg x tg 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
BAØI TAÄP 
π⎛ ⎞π⎜ ⎟⎝ ⎠,331. Tìm caùc nghieäm treân cuûa phöông trình: 
 π π⎛ ⎞ ⎛+ − = +⎜ ⎟ ⎜5 7sin 2x 3cos x 1 2sin x2 2 
⎞− ⎟⎝ ⎠ ⎝ ⎠
. Tìm caùc nghieäm x treân π⎛ ⎞⎜ ⎟2 ⎝ 2 ⎠0, cuûa phöông trình 
4x cos 6x sin 10,5 10x
3. Giaûi caùc phöông trình sau: 
x cos x 2 sin x s x+ = + 
 sin ( )− = π +2 2
 a/ sin co( )3 3 5 5
 b/ sin x sin2x sin3x 3
cos x cos2x cos3x
+ + =+ + 
 c/ 2 1 cos xtg x
1 sin x
+= − 
 d/ tg2x tg3x tg5x tg2x.tg3x.tg5x− − = 
 e/ 24cos x cos x
3
= 
 f/ 1 12 2 sin x
4 sin x cos x
π⎛ ⎞+ = +⎜ ⎟⎝ ⎠ 
2
 i/ 2tgx cot g2x 3
sin2x
+ = + 
 h/ 23tg3x cot g2x 2tgx
sin4x
+ = + 
 k/ =2 2 2sin x sin 2x sin 3x 2+ + 
 l/ 
si 2n x 2cos x 0
in x
+
1 s
=+ 
 m/ ( )225 4x 3sin2 x 8sin x 0− π + π =
 n/ sin x.cot g5x 1
cos9x
= 
 o/ 23tg6x 2tg2x cot g4x
sin8x
− = − 
 p/ ( )22sin 3x 1 4sin x 1− = 
 q/ 2 1 cos xtg +x
1 sin x
= − 
 r/ 3c x 33 2os cos3x sin xsin x
4
+ = 
 s/ 4 4x x 5sin cos
3 3 8
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 
 t/ =3 3 2cos x 4sin x 3cos xsin x sin x 0− − + 
 u/ 4 4x xsin cos 1 2sin x
2 2
+ = − 
 v/ s 3in x sin2x.sin x
4 4
π π⎛ ⎞ ⎛ ⎞− = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 w/ 
( )24
4
2 sin x sin3x
tg x 1
cos x
−+ = 
2 x y/ tgx cos x cos+ − x sin x 1 tg tgx
2
⎛ ⎞= + ⎠ 
. 
⎜ ⎟⎝
4 Cho phöông trình: ( ) ( ) (2 )2s x m 3 4cos x 1+ + = − 
 a/ Giaûi phöông trình khi m = 1 
2sin x 1 2cos2x in−
[ ]0,π b/ Tìm m ñeå (1) coù ñuùng 2 nghieäm treân 
m 0 m 1 m 3= ∨ ) 
5. Cho phöông trình: 
( )54 cos xs 5 2in x 4sin x.cos x sin 4x m 1− = + 
Bieát raèng x = π laø moät nghieäm cuûa (1). Haõy giaûi phöông trình trong tröôøng 
hôïp ñoù. 
Th.S Phạm Hồng Danh 
TT luyện thi Đại học CLC Vĩnh Viễn