1. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Ch動董ng I: PH働NG TRNH L働畛NG GIC
A. C S畛 L THUY畉T
1. Cung li棚n k畉t
a) Cung 畛i: ( ) ( )cos cos ; sin sin ;x x x x = =
b) Cung b湛: ( ) ( )cos cos ; sin sin ;x x x x = =
c) Cung ph畛: cos sin ; sin cos ; tan( ) cot ; cot tan
2 2 2 2
x x x x x x x x
錚 錚 錚 錚 錚 錚
= = = =錚 歎 錚 歎 錚 歎
錚 錚 錚 錚 錚 錚
d) Cung h董n k辿m : ( ) ( )cos cos ; sin sin ;x x x x + = + =
e) Cung h董n k辿m
2
: cos sin ; sin cos ;
2 2
x x x x
錚 錚 錚 錚
+ = + =錚 歎 錚 歎
錚 錚 錚 錚
2. C担ng th畛c l動畛ng gi叩c
a) C担ng th畛c c畛ng: b) C担ng th畛c nh但n 担i
( )cos cos cos sin sin
sin( ) sin cos cos sin
tan tan
tan( )
1 tan tan
cotacot 1
cot( )
cota cot
a b a b a b
a b a b a b
a b
a b
a b
b
a b
b
+ =
+ = +
+
+ =
+ =
+
2 2
2
2
2
sin2 2sin .cos
cos2 cos sin
2cos 1
1 2sin
2tan
tan2
1 tan
a a a
a a a
a
a
a
a
a
=
=
=
=
=
c) C担ng th畛c nh但n ba d) C担ng th畛c h畉 b畉c
3
3
sin3 3sin 4sin
cos3 4cos 3cos
a a a
a a a
=
=
2 2
3 3
1 cos2 1 cos2
sin ; cos
2 2
3sin sin3 3cos cos3
sin ; cos
4 4
a a
a a
a a a a
a a
+
= =
+
= =
e) C担ng th畛c t鱈ch thnh t畛ng f) C担ng th畛c t畛ng thnh t鱈ch
[ ]
[ ]
[ ]
1
cos cos cos( ) cos( )
2
1
sin sin cos( ) cos( )
2
1
sin cos sin( ) sin( )
2
a b a b a b
a b a b a b
a b a b a b
= + +
= +
= + +
cos cos 2cos cos
2 2
cos cos 2sin sin
2 2
sin sin 2sin cos
2 2
sin sin 2cos sin
2 2
a b a b
a b
a b a b
a b
a b a b
a b
a b a b
a b
+
+ =
+
=
+
+ =
+
=
3. H畉ng 畉ng th畛c th動畛ng d湛ng
( )
2 2 4 4 2 6 6 2
22 2
2 2
1 3
sin cos 1 sin cos 1 sin 2a sin cos 1 sin 2
2 4
1 1
1 tan 1+cot 1 sin2 sin cos
cos sin
a a a a a a a
a a a a a
a a
+ = + = + =
+ = = 賊 = 賊
4. Ph動董ng tr狸nh l動畛ng gi叩c c董 b畉n
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2. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
khi 1
2
sin ( ) ; sin sin( ) arcsin 2
2khi 1
( ) arcsin 2
VN m
x k
f x m xf x m k
x km
f x m k
留
留
留
>錚
= +錚錚
= = = +錚駕 錚 = + 錚逸４錚 = +錚逸３
khi 1
2
cos ( ) ; cos cos( ) arccos 2
2khi 1
( ) arccos 2
VN m
x k
f x m xf x m k
x km
f x m k
留
留
留
>錚
= +錚錚
= = = +錚駕 錚 = + 錚逸４錚 = +錚逸３
tan ( ) ( ) arctan ; tan tanf x m f x m k x x k 留 留 = = + = = +
cot ( ) ( ) arccot ; cot cotf x m f x m k x x k 留 留 = = + = = +
5. Ph動董ng tr狸nh th動畛ng g畉p
a. Ph動董ng tr狸nh b畉c 2
2 2 2
2 2 2
2
2
.sin ( ) .cos ( ) 0 sin ( ) 1 cos ( )
.cos ( ) .sin ( ) 0 ( ) 1 sin ( )
cos2 ( ) cos ( ) 0 cos2 ( ) 2cos ( ) 1
cos2 ( ) sin ( ) 0 cos2 ( ) 1 2sin ( )
.t
a f x b f x c Thay f x f x
a f x b f x c Thay f x f x
a f x b f x c Thay f x f x
a f x b f x c Thay f x f x
a
+ + = =
+ + = =
+ + = =
+ + = =
cos
1
an ( ) cot ( ) 0 cot ( )
tan ( )
f x b f x c Thay f x
f x
+ + = =
b. Ph動董ng tr狸nh d畉ng sin ( ) cos ( )a f x b f x c+ =
i畛u ki畛n c坦 nghi畛m: 2 2 2
a b c+
Chia 2 v畉 cho 2 2
a b+ , d湛ng c担ng th畛c c畛ng chuy畛n v畛 d畉ng c董 b畉n theo sin ho畉c cos.
c. Ph動董ng tr狸nh 畉ng c畉p
D畉ng 2 2
.sin .sin cos .cosa x b x x c x d+ + =
X辿t cosx = 0 c坦 th畛a m達n ph動董ng tr狸nh hay kh担ng.
X辿t cosx 0, chia 2 v畉 cho cos2
x 畛 動畛c ph動董ng tr狸nh b畉c 2 theo tanx.
C坦 th畛 thay v狸 x辿t cosx, ta c坦 th畛 thay b畉ng vi畛c x辿t sinx.
D畉ng 3 2 2 3
.sin .sin cos .sin .cos .cos 0a x b x x c x x d x+ + + =
X辿t cosx = 0 c坦 th畛a m達n ph動董ng tr狸nh hay kh担ng.
X辿t cosx 0, chia 2 v畉 cho cos3
x 畛 動畛c ph動董ng tr狸nh b畉c 3 theo tanx.
C坦 th畛 thay v狸 x辿t cosx, ta c坦 th畛 thay b畉ng vi畛c x辿t sinx.
d. Ph動董ng tr狸nh 畛i x畛ng lo畉i 1: (sin cos ) .sin cosa x x b x x c賊 + =
畉t t = sinx 賊 cosx, i畛u ki畛n 2t
Thay vo ph動董ng tr狸nh ta 動畛c ph動董ng tr狸nh b畉c 2 theo t.
e. Ph動董ng tr狸nh 畛i x畛ng lo畉i 2 : ( )tan cot ) (tan cot 0n n
a x x b x x+ + 賊 =
畉t t = tanx - cotx th狸 t R ; 畉t t = tanx + cotx th狸 2t .
Chuy畛n v畛 ph動董ng tr狸nh theo 畉n t.
f. C叩c ph動董ng ph叩p gi畉i ph動董ng tr狸nh l動畛ng gi叩c t畛ng qu叩t
Ph動董ng ph叩p bi畉n 畛i t動董ng 動董ng 動a v畛 d畉ng c董 b畉n
Ph動董ng ph叩p bi畉n 畛i ph動董ng tr狸nh 達 cho v畛 d畉ng t鱈ch.
Ph動董ng ph叩p 畉t 畉n ph畛.
Ph動董ng ph叩p 畛i l畉p.
Ph動董ng ph叩p t畛ng b狸nh ph動董ng.
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3. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
B. BI T畉P LUY畛N T畉P
D畉ng 1 : Ph動董ng tr狸nh l動畛ng gi叩c c董 b畉n.
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. cos sin2 0
3
x x
錚 錚
+ + =錚 歎
錚 錚
2. cos cos 1
3 3
x x
錚 錚 錚 錚
+ + =錚 歎 錚 歎
錚 錚 錚 錚
3. tan2 .tan 1x x =
4. 2 2 2
sin sin .tan 3x x x+ = 5. 2 2
5cos sin 4x x+ = 3.
1
3sin cos
cos
x x
x
+ =
7. 4 4
cos 2 sin3 sin 2x x x= 8. tan 1 tan
4
x x
錚 錚
= 錚 歎
錚 錚
9.
3 31
sin cos cos sin
4
x x x x= +
10. 4 4
sin cos cos4x x x+ = 11. cos7x - sin5x = ( cos5x - sin7x) 12. sin + cos =
13. 2 2
sin 5 cos 3 1x x+ = 14.
2
cos cos2 cos4
16
x x x
= 15. ( )sin sin 1x =
16.
2 2
cos sin
1 sin 1 cos
x x
x x
=
17.
1 1 2
cos sin2 sin4x x x
+ = 18. 3 2
4sin 2 6sin 3x x+ =
Bi 2 : Cho ph動董ng tr狸nh ( ) ( )tan cos cot sinx x =
1. T狸m i畛u ki畛n x叩c 畛nh c畛a ph動董ng tr狸nh.
2. T狸m t畉t c畉 c叩c nghi畛m thu畛c o畉n [ ]3 ; c畛a ph動董ng tr狸nh.
Bi 3 : Cho ph動董ng tr狸nh sin6
x + cos6
x = m.
1. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 炭ng 2 nghi畛m trong kho畉ng ( )0;
Bi 4: Gi畉i v bi畛n lu畉n ph動董ng tr狸nh ( ) 2
2 1 cos2 2 sin 3 2 0m x m x m + + =
D畉ng 2 : Ph動董ng tr狸nh b畉c nh畉t, b畉c hai.
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1.
2
2cos 5sin 4 0
3 3
x x
錚 錚 錚 錚
+ + + =錚 歎 錚 歎
錚 錚 錚 錚
2.
5
cos2 4cos 0
2
x x + =
3. 4 4
sin cos cos2x x x+ = 4. 4 4 1
cos sin sin2
2
x x x+ =
5. ( )2
2 2cos 3 2 2 cos3 1 0x x + + = 6. 4 4
cos sin 2sin 1
2 2
x x
x+ + =
7. ( )6 6
4 sin cos cos 2 0
2
x x x
錚 錚
+ =錚 歎
錚 錚
8. 2tan 3cot 4x x+ =
9. 4 2 1
cos sin
4
x x= 10.
2 2
6 6
cos sin
4cot2
sin cos
x x
x
x x
=
+
11.
1
2tan cot 2sin2
sin2
x x x
x
+ = + 12.
8 8 217
sin cos cos 2
16
x x x+ =
13. 4cos cos4 1 2cos2x x x = + 14. 5 5 2
4sin cos 4cos sin cos 4 1x x x x x = +
15. 2 2
cos4 cos 3 cos 1x x x= + 16. sin3 cos2 1 2sin cos2x x x x+ = +
Bi 2 : Cho ph動董ng tr狸nh sin3 cos2 ( 1)sin 0x m x m x m + + =
1. Gi畉i ph動董ng tr狸nh khi m = 2.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 炭ng 4 nghi畛m thu畛c kho畉ng ( )0;2
D畉ng 3 : Ph動董ng tr狸nh b畉c nh畉t theo sinx, cosx.
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4. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. 3sin cos 2 0x x + = 2. 3
3sin 1 4sin 3cos3x x x = +
3.
4 4
sin cos 1
4
x x
錚 錚
+ + =錚 歎
錚 錚
4. ( )4 4
2 cos sin 3sin4 2x x x+ + =
5. 2sin2 2sin4 0x x+ = 6. 3sin2 2cos2 3x x+ =
7.
9
3cos 2 3sin
2
x x+ = 8. 4cos3 3sin3 5 0x x + =
9. 2
sin cos sin cos2x x x x = 10. ( )tan 3cot 4 sin 3cosx x x x = +
11. 2sin3 3cos7 sin7 0x x x+ + = 12. ( )cos5 sin3 3 cos3 sin5x x x x =
13. ( ) ( ) 2
2sin cos 1 cos sinx x x x + = 14. 1 cos sin3 cos3 sin2 sinx x x x x+ + =
15. 3
3sin 1 4sin 3cos3x x x = + 16. 3sin cos 2cos 2
3
x x x
錚 錚
+ + =錚 歎
錚 錚
Bi 2 : Cho ph動董ng tr狸nh ( )3 sin 2 1 cos 3 1m x m x m+ = +
1. Gi畉i ph動董ng tr狸nh khi m = 1.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m.
Bi 3 : T狸m gi叩 tr畛 l畛n nh畉t, gi叩 tr畛 nh畛 nh畉t c畛a hm s畛
1.
cos sin 1
sin 2cos 4
x x
y
x x
+
=
+
2.
cos3 sin3 1
cos3 2
x x
y
x
+ +
=
+
3.
1 3sin 2cos
2 sin cos
x x
y
x x
+
=
+ +
4.
2
sin cos cos
sin cos 1
x x x
y
x x
+
=
+
D畉ng 4 : Ph動董ng tr狸nh 畉ng c畉p
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. 2 2
2sin sin cos 3cos 0x x x x+ = 2. 2
2sin2 3cos 5sin cos 2 0x x x x + =
3. 2 2
sin sin2 2cos 0,5x x x+ = 4. 2
sin2 2sin 2cos2x x x =
5. 2sin2
x + 3sinx.cosx - 3cos2
x = 1 6. 2 21
4 3 3
2 2 2
os sin sin
x x
c x+ + =
7. ( )2 2
3sin 4sin 2 8 3 9 cos 0+ + =x x x 8. 3 3
2cos 3cos 8sin 0x x x+ =
9.
3 3 8
3cos 5sin 7sin cos 0
3
x x x x + = 10.
3 5sin4 cos
6sin 2cos
2cos2
x x
x x
x
=
11.
2
sin 2sin
4
x x
錚 錚
+ =錚 歎
錚 錚
12. 3 2cos sin cos3 3 2sin sin2x x x x x = +
13. 2 2
3sin 2sin2 cos 0x x x + = 14.
3
12sin 2sin
4
x x
錚 錚
=錚 歎
錚 錚
Bi 2 : Cho ph動董ng tr狸nh ( ) ( )2 2
sin 3 sin2 2 cos 0m x m x m x + =
1. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m duy nh畉t thu畛c kho畉ng 0,
4
錚 錚
錚 歎
錚 錚
.
D畉ng 5 : Ph動董ng tr狸nh 畛i x畛ng lo畉i 1
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5. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. ( )2 sin cos sin2 1 0x x x+ + + = 2. ( )sin cos 6 sin cos 1x x x x=
3. sin2 2sin 1
4
x x
錚 錚
+ =錚 歎
錚 錚
4. tan 2 2sin 1x x =
5. 3 3
sin cos 1x x+ = 6. ( ) ( )1 sin 1 cos 2+ + =x x
7. 2sin tan cot
4
錚 錚
+ = +錚 歎
錚 錚
x x x
p
8. ( )
3
sin cos sin cos 1 0x x x x+ + =
9. ( )
4
sin cos 3sin2 1 0x x x+ = 10. 3 3
cos sin cos2x x x =
11. ( )3 3
sin cos 2 sin cos 3sin2 0x x x x x+ + + = 12. ( )
3
sin cos 1 sin cosx x x x = +
13.
1 1
sin cos 2 tan cot 0
sin cos
x x x x
x x
+ + + + + + = 14. ( ) ( )1 sin2 sin cos cos2x x x x + =
Bi 2 : Cho ph動董ng tr狸nh 3 3
cos sinx x m = . X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m.
D畉ng 5 : Ph動董ng tr狸nh 畛i x畛ng lo畉i 2
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. ( ) ( )2 2
3 tan cot 2 tan cot 2 0x x x x+ + = 2. 7 7
tan cot tan cotx x x x+ = +
3. 2 3 2 3
tan tan tan cot cot cot 6x x x x x x+ + + + + = 4. ( ) ( )4 2 2
9 tan cot 48 tan cot 96x x x x+ = + +
5. ( ) 2 2
3 tan cot tan cot 6x x x x + + = 6. ( ) ( )4 2 2
3 tan cot 8 tan cot 21+ + =x x x x
Bi 2 : Cho ph動董ng tr狸nh ( ) ( )2 2 2
tan cot 2 2 tan cotx x m x x m m+ + + + = . X叩c 畛nh m 畛 ph動董ng
tr狸nh c坦 nghi畛m.
D畉ng 6 : Bi畉n 畛i t動董ng 動董ng d動a v畛 d畉ng c董 b畉n
Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. 3 3 3
sin cos sin cos
8
x x x x = 2. 2 2 2 2
cos cos 2 cos 3 cos 4 2x x x x+ + + =
3. ( )3 3 5 5
sin cos 2 in cosx x s x x+ = + 4. ( )8 8 10 10 5
sin cos 2 sin cos cos2
4
x x x x x+ = + +
5.
sin cot5
1
cot
x x
x
= 6. 6tan 5cot3 tan 2+ =x x x
D畉ng 7 : Bi畉n 畛i bi畉n 畛i t鱈ch b畉ng 0
1/ cos2x- cos8x+ cos4x=1 2/sinx+2cosx+cos2x-2sinxcosx=0
3/sin2x-cos2x=3sinx+cosx-2 4/sin
3
x+2cosx-2+sin
2
x=0
5/ 3sinx+2cosx=2+3tanx 6/
3
2
sin2x+ 2 cos
2
x+ 6 cosx=0
7/ 2sin2x-cos2x=7sinx+2cosx-4 8/
sin3 sin5
3 5
x x
=
9/ 2cos2x-8cosx+7=
1
cos x 10/ cos
8
x+sin
8
x=2(cos
10
x+sin
10
x)+
5
4
cos2x
11/ 1+ sinx+ cos3x= cosx+ sin2x+ cos2x 12/ 1+sinx+cosx+sin2x+cos2x=0
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6. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
13/ sin
2
x(tanx+1)=3sinx(cosx-sinx)+3 14/ 2sin3x-
1
sin x =2cos3x+
1
cos x
15/cos
3
x+cos
2
x+2sinx-2=0 16/cos2x-2cos
3
x+sinx=0
17/ tanxsin2x-cos2x+2(2cosx-
1
cos x
)=0 18/sin2x=1+ 2 cosx+cos2x
D畉ng 7 : Bi畉n 畛i bi畉n 畛i t鱈ch thnh t畛ng, ho畉c t畛ng thnh t鱈ch
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. sinx + sin2x + sin3x = cosx + cos 2x + cos3x 2. sin
2
x + sin
2
2x = sin
2
3x + sin
2
4x
3. sin
2
x + sin
2
2x + sin
2
3x + sin
2
4x = 2 4.
2 2 2 3
cos cos 2 cos 3
2
x x x+ + =
5. sin5x.cos6x+ sinx = sin7x.cos4x 6.
1
sin sin
3 3 2
x x
錚 錚 錚 錚
+ =錚 歎 錚 歎
錚 錚 錚 錚
7.
1
sin cos
4 12 2
x x
錚 錚 錚 錚
+ + =錚 歎 錚 歎
錚 錚 錚 錚 8. cosx. cos4x - cos5x=0
9. sin6x.sin2x = sin5x.sin3x 10. 2 + sinx.sin3x = 2 cos 2x
Bi 2 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1/ sin
2
x+sin
2
3x=cos
2
2x+cos
2
4x 2/ cos
2
x+cos
2
2x+cos
2
3x+cos
2
4x=3/2
3/sin
2
x+ sin
2
3x-3 cos
2
2x=0 4/ cos3x+ sin7x=2sin
2
(
5
4 2
x
+ )-2cos
2 9
2
x
5/ sin
2
4 x+ sin
2
3x= cos
2
2x+ cos
2
x 6/sin
2
4x-cos
2
6x=sin(10,5 10x + )
7/ cos
4
x-5sin
4
x=1 8/4sin
3
x-1=3- 3 cos3x
9/ sin
2
2x+ sin
2
4x= sin
2
6x 10/ sin
2
x= cos
2
2x+ cos
2
3x
11/ 4sin
3
xcos3x+4cos
3
x sin3x+3 3 cos4x=3 12/ 2cos
2
2x+ cos2x=4 sin
2
2xcos
2
x
D畉ng 8 : 畉t 畉n ph畛
Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. tan2 2tan sin2 0x x x + = 2.
2 2
cos 2 cos cos 2 cos 3x x x x+ + = 3.
5
3sin cos 3
3sin cos 3
x x
x x
+ + =
+ +
4.
2
cos 2 2 cos 2x x+ + =
D畉ng 9 : Ph動董ng ph叩p 畛i l畉p
Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. 3 4
sin cos 1x x+ = 2. 2010 2010
sin cos 1x x+ =
3. 2 2
3cos 1 sin 7x x+ = 4. sin3 .cos4 1x x =
5. 3 3 2
sin cos 2 sin 2x x x+ = 6. cos2 .cos5 1x x =
D畉ng 10 : Ph動董ng ph叩p t畛ng b狸nh ph動董ng
Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. ( )3
cos2 cos6 4 3sin 4sin 1 0x x x x + + = 2. 2
3sin2 2sin 4cos 6 0x x x + =
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7. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
3. 2sin2 cos2 2 2sin 4 0x x x+ + = 4. 2
cos2 3sin2 4sin 2sin 4 2 3cosx x x x x + + =
C. BI T畉P T畛NG H畛P
Bi 1 2 2
cos 3sin 2 1 sinx x x = +
Bi 2 3 3 2
cos 4sin 3cos .sin sin 0x x x x x + =
Bi 3 Gi畉i ph動董ng tr狸nh: sin 2 2tan 3x x+ =
3
sin .sin 2 sin3 6cosx x x x+ =
Bi 4
2cos2 1
cot 1 sin sin 2
1 tan 2
x
x x x
x
= +
+
Bi 5 sin3 cos3 2cos 0x x x+ + =
Bi 6 3
sin 4sin cos 0x x x + =
Bi 7 2 2
tan .sin 2sin 3(cos2 sin cos )x x x x x x = +
Bi 8 cos3 4cos2 3cos 4 0x x x + =
Bi 9 (2cos 1)(2sin cos ) sin 2 sinx x x x x + =
Bi 10 cos cos2 cos3 cos4 0x x x x+ + + =
Bi 11 2 2 2 2
sin sin 3 cos 2 cos 4x x x x+ = +
Bi 12 3 3 3
sin cos3 cos sin3 sin 4x x x x x+ =
Bi 13 3 3 2
4sin 3cos 3sin sin cos 0x x x x x+ =
Bi 14 Gi畉i ph動董ng tr狸nh:
2
(2sin 1)(3cos4 2sin 4) 4cos 3x x x x+ + + =
Bi 15 6 6 8 8
sin cos 2(sin cos )x x x x+ = +
Bi 16
1
cos .cos2 .cos4 .cos8
16
x x x x =
Bi 17
3
8cos cos3
3
x x
錚 錚
+ =錚 歎
錚 錚
Bi 18 Gi畉i ph動董ng tr狸nh:
2
(2sin 1)(2sin 2 1) 3 4cosx x x + =
Bi 19 Gi畉i ph動董ng tr狸nh:
cos2 cos8 cos6 1x x x + =
Bi 20 Gi畉i ph動董ng tr狸nh:
sin 4 4sin 4cos cos4 1x x x x + =
Bi 21 Gi畉i ph動董ng tr狸nh:
3sin 2cos 2 3tanx x x+ = +
Bi 22 Gi畉i ph動董ng tr狸nh:
3
2cos cos2 sin 0x x x+ + =
Bi 23 Gi畉i ph動董ng tr狸nh:
2(tan sin ) 3(cot cos ) 5 0x x x x + + =
Bi 24 Gi畉i ph動董ng tr狸nh:
4cos 2cos2 cos4 1x x x =
Bi 25 Gi畉i ph動董ng tr狸nh:
sin sin 2 sin3
3
cos cos2 cos3
x x x
x x x
+ +
=
+ +
Bi 26 Gi畉i ph動董ng tr狸nh:
sin .sin 4 2cos 3 cos .sin 4
6
x x x x x
錚 錚
= 錚 歎
錚 錚
Bi 27 Gi畉i ph動董ng tr狸nh:
2 2
1 sin sin cos sin 2 os
2 2 4 2
x x x
x x c
錚 錚
+ = 錚 歎
錚 錚
Bi 28 Gi畉i ph動董ng tr狸nh:
2cos2 sin 2 2(sin cos )x x x x = +
Bi 29 Gi畉i ph動董ng tr狸nh:
1
cos cos2 cos3
2
x x x + =
Bi 30 Gi畉i ph動董ng tr狸nh:
3
sin 2 sin
4
x x
錚 錚
+ =錚 歎
錚 錚
Bi 31 Gi畉i ph動董ng tr狸nh:
1 sin cos sin 2 cos2 0x x x x+ + + + =
Bi 32 Gi畉i ph動董ng tr狸nh:
2 3 2 3
tan tan tan 6x x x cotx cot x cot x+ + + + + =
Bi 33 Gi畉i ph動董ng tr狸nh: 1 sin3 sin cos2x x x+ = +
Bi 34 Gi畉i ph動董ng tr狸nh:
4 4 7
sin cos cot .cot
8 3 6
x x x x
錚 錚 錚 錚
+ = + 錚 歎 錚 歎
錚 錚 錚 錚
Bi 35 Gi畉i ph動董ng tr狸nh:
2 3
cos 2 2(sin cos ) 3sin 2 3 0x x x x+ + =
Bi 36 Gi畉i ph動董ng tr狸nh:
4(sin3 cos2 ) 5(sin 1)x x x =
Bi 37 Gi畉i ph動董ng tr狸nh: 3
sin 4sin cos 0x x x + =
Bi 38 Gi畉i ph動董ng tr狸nh:
3
cos10 1 cos8 6cos3 .cos cos 8cos .cos 3x x x x x x x+ + + = +
Bi 39 Gi畉i ph動董ng tr狸nh:
4 4 1
sin cos
4 4
x x
錚 錚
+ + =錚 歎
錚 錚
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8. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Bi 40 Gi畉i ph動董ng tr狸nh:
3 3 2
cos .cos3 sin .sin3
4
x x x x+ =
Bi 41 Gi畉i ph動董ng tr狸nh:
3 3 3 3
(sin sin 2 sin3 ) sin sin 2 sin 3x x x x x x+ + = + +
Bi 42 Gi畉i ph動董ng tr狸nh:
3 1
8sin
cos sin
x
x x
= +
D. GI畛I THI畛U 畛 THI TUY畛N SINH CC NM
A02:Tm no thu辿c (0;2 ) c単a PT:
5 3
錚 錚
錚 歎
錚 錚
+
+ = +
+
cosx sin3x
sinx cos2x
1 2sin2x
B02: GPT:
2 2 2 2sin 3x cos 4x sin 5x cos 6x. =
D02: Tm no thu辿c [0;14] c単a PT:
cos3 4cos2 3cos 4 0.x x x + =
A03: Gi其i ph測ng trnh:
cos2x 12
cot x 1 sin x sin 2x.
1 tan x 2
= +
+
B03: Gi其i ph測ng trnh:
2
cot x tan x 4 sin 2x .
sin 2x
+ =
D03: Gi其i ph測ng trnh
xx2 2 2
sin tan x cos 0.
22 4
錚 錚
=錚 歎
錚 錚
B04: Gi其i ph測ng trnh
( ) 2
5sin x 2 3 1 sin x tan x. =
D04: Gi其i ph測ng trnh
( )( )2cosx 1 2sin x cosx sin 2x sin x. + =
A-05: GPT: cos2
3x.cos2x-cos2
x = 0
A-06: GPT: ( )6 62 sin cos sin cos
0
2 2sin
x x x x
x
+
=
B-06: GPT: cot sin 1 tan tan 4
2
x
x x x
錚 錚
+ + =錚 歎
錚 錚
D-06: GPT: cos3x+cos2x-cosx-1=0
2 2
A07: GPT: (1 sin ) cos (1 cos )sin 1 sin 2
2B07: GPT: 2sin 2 sin7 1 sin
2
D07: GPT: sin cos 3cos 2
2 2
x x x x x
x x x
x x
x
+ + + = +
+ =
錚 錚
+ + =錚 歎
錚 錚
A08: GPT
1 1 7
4sin .
3sin 4
sin
2
x
x
x
+ =
錚 錚
錚 歎
錚 錚 錚 錚
錚 歎
錚 錚
B08: GPT
3 3 2 2sin 3 cos sin cos 3sin cos .x x x x x x =
D08: GPT
2sin (1 cos2 ) sin 2 1 2cos .x x x x+ + = +
A09: GPT
(1 2sin )cos
3
(1 2sin )(1 sinx)
x x
x
=
+
.
B09: GPT
3
sinx cos sin 2 3 os3 2( os4 sin ).x x c x c x x+ + = +
D09: GPT
3 os5 2sin3 cos2 sinx 0.c x x x =
A10: GPT
(1 sinx os2 )sin
14
cos .
1 t anx 2
c x x
x
錚 錚
+ + +錚 歎
錚 錚 =
+
B10: GPT
(sin2 os2 )cos 2cos2 sinx 0.x c x x x+ + =
D10: GPT
sin 2 os2 3sin cos 1 0.x c x x x + =
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9. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
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10. NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
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![NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Ch動董ng I: PH働NG TRNH L働畛NG GIC
A. C S畛 L THUY畉T
1. Cung li棚n k畉t
a) Cung 畛i: ( ) ( )cos cos ; sin sin ;x x x x = =
b) Cung b湛: ( ) ( )cos cos ; sin sin ;x x x x = =
c) Cung ph畛: cos sin ; sin cos ; tan( ) cot ; cot tan
2 2 2 2
x x x x x x x x
錚 錚 錚 錚 錚 錚
= = = =錚 歎 錚 歎 錚 歎
錚 錚 錚 錚 錚 錚
d) Cung h董n k辿m : ( ) ( )cos cos ; sin sin ;x x x x + = + =
e) Cung h董n k辿m
2
: cos sin ; sin cos ;
2 2
x x x x
錚 錚 錚 錚
+ = + =錚 歎 錚 歎
錚 錚 錚 錚
2. C担ng th畛c l動畛ng gi叩c
a) C担ng th畛c c畛ng: b) C担ng th畛c nh但n 担i
( )cos cos cos sin sin
sin( ) sin cos cos sin
tan tan
tan( )
1 tan tan
cotacot 1
cot( )
cota cot
a b a b a b
a b a b a b
a b
a b
a b
b
a b
b
+ =
+ = +
+
+ =
+ =
+
2 2
2
2
2
sin2 2sin .cos
cos2 cos sin
2cos 1
1 2sin
2tan
tan2
1 tan
a a a
a a a
a
a
a
a
a
=
=
=
=
=
c) C担ng th畛c nh但n ba d) C担ng th畛c h畉 b畉c
3
3
sin3 3sin 4sin
cos3 4cos 3cos
a a a
a a a
=
=
2 2
3 3
1 cos2 1 cos2
sin ; cos
2 2
3sin sin3 3cos cos3
sin ; cos
4 4
a a
a a
a a a a
a a
+
= =
+
= =
e) C担ng th畛c t鱈ch thnh t畛ng f) C担ng th畛c t畛ng thnh t鱈ch
[ ]
[ ]
[ ]
1
cos cos cos( ) cos( )
2
1
sin sin cos( ) cos( )
2
1
sin cos sin( ) sin( )
2
a b a b a b
a b a b a b
a b a b a b
= + +
= +
= + +
cos cos 2cos cos
2 2
cos cos 2sin sin
2 2
sin sin 2sin cos
2 2
sin sin 2cos sin
2 2
a b a b
a b
a b a b
a b
a b a b
a b
a b a b
a b
+
+ =
+
=
+
+ =
+
=
3. H畉ng 畉ng th畛c th動畛ng d湛ng
( )
2 2 4 4 2 6 6 2
22 2
2 2
1 3
sin cos 1 sin cos 1 sin 2a sin cos 1 sin 2
2 4
1 1
1 tan 1+cot 1 sin2 sin cos
cos sin
a a a a a a a
a a a a a
a a
+ = + = + =
+ = = 賊 = 賊
4. Ph動董ng tr狸nh l動畛ng gi叩c c董 b畉n
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![NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
B. BI T畉P LUY畛N T畉P
D畉ng 1 : Ph動董ng tr狸nh l動畛ng gi叩c c董 b畉n.
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1. cos sin2 0
3
x x
錚 錚
+ + =錚 歎
錚 錚
2. cos cos 1
3 3
x x
錚 錚 錚 錚
+ + =錚 歎 錚 歎
錚 錚 錚 錚
3. tan2 .tan 1x x =
4. 2 2 2
sin sin .tan 3x x x+ = 5. 2 2
5cos sin 4x x+ = 3.
1
3sin cos
cos
x x
x
+ =
7. 4 4
cos 2 sin3 sin 2x x x= 8. tan 1 tan
4
x x
錚 錚
= 錚 歎
錚 錚
9.
3 31
sin cos cos sin
4
x x x x= +
10. 4 4
sin cos cos4x x x+ = 11. cos7x - sin5x = ( cos5x - sin7x) 12. sin + cos =
13. 2 2
sin 5 cos 3 1x x+ = 14.
2
cos cos2 cos4
16
x x x
= 15. ( )sin sin 1x =
16.
2 2
cos sin
1 sin 1 cos
x x
x x
=
17.
1 1 2
cos sin2 sin4x x x
+ = 18. 3 2
4sin 2 6sin 3x x+ =
Bi 2 : Cho ph動董ng tr狸nh ( ) ( )tan cos cot sinx x =
1. T狸m i畛u ki畛n x叩c 畛nh c畛a ph動董ng tr狸nh.
2. T狸m t畉t c畉 c叩c nghi畛m thu畛c o畉n [ ]3 ; c畛a ph動董ng tr狸nh.
Bi 3 : Cho ph動董ng tr狸nh sin6
x + cos6
x = m.
1. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 nghi畛m.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 炭ng 2 nghi畛m trong kho畉ng ( )0;
Bi 4: Gi畉i v bi畛n lu畉n ph動董ng tr狸nh ( ) 2
2 1 cos2 2 sin 3 2 0m x m x m + + =
D畉ng 2 : Ph動董ng tr狸nh b畉c nh畉t, b畉c hai.
Bi 1 : Gi畉i c叩c ph動董ng tr狸nh l動畛ng gi叩c sau :
1.
2
2cos 5sin 4 0
3 3
x x
錚 錚 錚 錚
+ + + =錚 歎 錚 歎
錚 錚 錚 錚
2.
5
cos2 4cos 0
2
x x + =
3. 4 4
sin cos cos2x x x+ = 4. 4 4 1
cos sin sin2
2
x x x+ =
5. ( )2
2 2cos 3 2 2 cos3 1 0x x + + = 6. 4 4
cos sin 2sin 1
2 2
x x
x+ + =
7. ( )6 6
4 sin cos cos 2 0
2
x x x
錚 錚
+ =錚 歎
錚 錚
8. 2tan 3cot 4x x+ =
9. 4 2 1
cos sin
4
x x= 10.
2 2
6 6
cos sin
4cot2
sin cos
x x
x
x x
=
+
11.
1
2tan cot 2sin2
sin2
x x x
x
+ = + 12.
8 8 217
sin cos cos 2
16
x x x+ =
13. 4cos cos4 1 2cos2x x x = + 14. 5 5 2
4sin cos 4cos sin cos 4 1x x x x x = +
15. 2 2
cos4 cos 3 cos 1x x x= + 16. sin3 cos2 1 2sin cos2x x x x+ = +
Bi 2 : Cho ph動董ng tr狸nh sin3 cos2 ( 1)sin 0x m x m x m + + =
1. Gi畉i ph動董ng tr狸nh khi m = 2.
2. X叩c 畛nh m 畛 ph動董ng tr狸nh c坦 炭ng 4 nghi畛m thu畛c kho畉ng ( )0;2
D畉ng 3 : Ph動董ng tr狸nh b畉c nh畉t theo sinx, cosx.
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![NGUY畛N T畉N TI THPT LAI VUNG I 畛NG THP
Bi 40 Gi畉i ph動董ng tr狸nh:
3 3 2
cos .cos3 sin .sin3
4
x x x x+ =
Bi 41 Gi畉i ph動董ng tr狸nh:
3 3 3 3
(sin sin 2 sin3 ) sin sin 2 sin 3x x x x x x+ + = + +
Bi 42 Gi畉i ph動董ng tr狸nh:
3 1
8sin
cos sin
x
x x
= +
D. GI畛I THI畛U 畛 THI TUY畛N SINH CC NM
A02:Tm no thu辿c (0;2 ) c単a PT:
5 3
錚 錚
錚 歎
錚 錚
+
+ = +
+
cosx sin3x
sinx cos2x
1 2sin2x
B02: GPT:
2 2 2 2sin 3x cos 4x sin 5x cos 6x. =
D02: Tm no thu辿c [0;14] c単a PT:
cos3 4cos2 3cos 4 0.x x x + =
A03: Gi其i ph測ng trnh:
cos2x 12
cot x 1 sin x sin 2x.
1 tan x 2
= +
+
B03: Gi其i ph測ng trnh:
2
cot x tan x 4 sin 2x .
sin 2x
+ =
D03: Gi其i ph測ng trnh
xx2 2 2
sin tan x cos 0.
22 4
錚 錚
=錚 歎
錚 錚
B04: Gi其i ph測ng trnh
( ) 2
5sin x 2 3 1 sin x tan x. =
D04: Gi其i ph測ng trnh
( )( )2cosx 1 2sin x cosx sin 2x sin x. + =
A-05: GPT: cos2
3x.cos2x-cos2
x = 0
A-06: GPT: ( )6 62 sin cos sin cos
0
2 2sin
x x x x
x
+
=
B-06: GPT: cot sin 1 tan tan 4
2
x
x x x
錚 錚
+ + =錚 歎
錚 錚
D-06: GPT: cos3x+cos2x-cosx-1=0
2 2
A07: GPT: (1 sin ) cos (1 cos )sin 1 sin 2
2B07: GPT: 2sin 2 sin7 1 sin
2
D07: GPT: sin cos 3cos 2
2 2
x x x x x
x x x
x x
x
+ + + = +
+ =
錚 錚
+ + =錚 歎
錚 錚
A08: GPT
1 1 7
4sin .
3sin 4
sin
2
x
x
x
+ =
錚 錚
錚 歎
錚 錚 錚 錚
錚 歎
錚 錚
B08: GPT
3 3 2 2sin 3 cos sin cos 3sin cos .x x x x x x =
D08: GPT
2sin (1 cos2 ) sin 2 1 2cos .x x x x+ + = +
A09: GPT
(1 2sin )cos
3
(1 2sin )(1 sinx)
x x
x
=
+
.
B09: GPT
3
sinx cos sin 2 3 os3 2( os4 sin ).x x c x c x x+ + = +
D09: GPT
3 os5 2sin3 cos2 sinx 0.c x x x =
A10: GPT
(1 sinx os2 )sin
14
cos .
1 t anx 2
c x x
x
錚 錚
+ + +錚 歎
錚 錚 =
+
B10: GPT
(sin2 os2 )cos 2cos2 sinx 0.x c x x x+ + =
D10: GPT
sin 2 os2 3sin cos 1 0.x c x x x + =
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