Nguyen Tat Mao Sinh Vien dai Hoc Hang Hai Viet Nam
C¸c d¹ng bt ph ¬ng tr×nh l îng gi¸c
Lo¹i 1. Bi n lu n theo kệ ậ
1. sin (πcosx) = 1
2. cos(8sinx) = -1
3. tan(πcosx ) = cot(π sinx)
4. cos(πsinx) = cos(3πsinx)
5. tan(π cosx) = tan(2π cosx)
6. sinx2 =
1
2
8. cot(x2 + 4x + 3) = cot6
9. Tìm nghi m d ng nh nh t c a pt ệ ươ ỏ ấ ủ
cos
22
)1(cos += xx
ππ
10. Tìm nghi m d ng nh nh t c a pt ệ ươ ỏ ấ ủ
sin
)2(sin
22
xxx
+=
ππ
11. Tìm nghi m d ng nh nh t c a pt ệ ươ ỏ ấ ủ
cos
0sin)2/12(
22
=−−+
xxx
ππ
Lo¹i 2. Công th c h b cứ ạ ậ
1. 4cos2(2x - 1) = 1
2. 2sin2 (x + 1) = 1
3. cos2 3x + sin2 4x = 1
4. sin(1 - x) =
2
3
5. 2cosx + 1 = 0
6. tan2 (2x –
3
π
) = 2
7. cos2 (x –
5
π
) = sin2(2x +
4
5
π
)
Lo¹i 3. Công th c c ng, bi n đ iứ ộ ế ổ
1. sin2x + cos2x =
2
sin3x
2. cos3x – sinx =
3
(cosx –sin3x )
3.
05cos
2
1
5sin
2
3
)3
2
cos( =++− xxx
π
4. sin3x =
2
cos(x – π /5) + cos3x
5. sin(x + π /4) + cos(x + π /4) =
2
cos7x
6. Tìm t t c các nghi m xấ ả ệ
);
2
3
(
π
π
−∈
c a pt: sinxcosủ
8
π
+ cosxsin
8
π
=
1
2
Lo¹i 4. Bài toán bi n lu n theo mệ ậ
1. Gi i và bi n lu n ả ệ ậ
2sin(1-2x) = m
2. 3cos23x = m
3. sin3x + cos3x = m
4. m.sin2 2x + cos4x = m
5. Gi i và bi n lu n ả ệ ậ
sin2x – 2m = (6m + 7)sin2x
6. Gi i và bi n lu n ả ệ ậ
(3m + 5).sin(x + π/2) = (2m + 3)cosx -m
7. Gi i và bi n lu n ả ệ ậ
cos3x + m – 5 = (3- 2m)cos3x
8. Cho pt sin4x + cos4x = m
a) Xác đ nh m đ pt có nghi mị ể ệ
b) Gi i pt v i m = ¾ả ớ
Lo¹i 5. T ng h pổ ợ
1. cos22x – sin28x = sin(
x10
2
17 +
π
)
2. sin23x – cos24x = sin25x – cos26x
3.
x
x
xcos2
sin1
2sin
−=
+
4.
xxx 4sin
2
2sin
1
cos
1
=+
5. Tìm t t c các nghi m xấ ả ệ
)3;
2
(
π
π
∈
c a pt:ủ
sin(2x +
)
2
7
cos(3)
2
5
ππ
−−
x
= 1 + 2sinx
6. Gi i pt:ả
4sin3xcos3x +4cos3xsin3x + 3
3
cos4x = 3
7.
)
8
(cos2)
8
cos()
8
sin(32
2
πππ
−+−−
xxx
=
x))
3
x)cos(-
3
cos(x(sin43
2
+++
ππ
8. 4sin32x + 6sin2x = 3
9. Tìm nghi m nguyên c a pt: ệ ủ
1
Nguyen Tat Mao Sinh Vien dai Hoc Hang Hai Viet Nam
1)80016093(
8
cos
2
=
++−
xxx
π
D¹ng 2: Ph ¬ng tr×nh bËc nhÊt, bËc hai vµ bËc cao ®èi víi mét hµm sè
l îng gi¸c
1/
2cos2x - 4cosx =1
sinx 0
≥
2/ 4sin3x + 3
2
sin2x = 8sinx
3/ 4cosx.cos2x + 1 = 0 4/
1-5sinx + 2cosx = 0
cosx 0
≥
5/ Cho 3sin3x - 3cos2x + 4sinx - cos2x + 2 = 0(1) vµ cos2x +
3cosx(sin2x - 8sinx) = 0(2)
T×m n0 cña (1) ®ång thêi lµ n0 cña (2) ( nghiÖm chung sinx =
1
3
)
6/ sin3x + 2cos2x - 2 = 0 7/ tanx +
3
cotx
- 2 = 0
b /
2
4
cos x
+ tanx = 7 c / sin6x + cos4x = cos2x
8/ sin(
5π
2x + 2
) - 3cos(
7
2
x
π
−
) = 1 + 2sinx
9/
2
sin x - 2sinx + 2 = 2sinx -1
10/ cos2x + 5sinx + 2 = 0
11/ tanx + cotx = 4 12/
2 4
sin 2x + 4cos 2x -1 = 0
2sinxcosx
13/
sin 1 cos 0x x+ + =
14/ cos2x + 3cosx + 2 = 0
15/
2 4
4sin 2 6sin 9 3cos 2 0
cos
x x x
x
+ − − =
16/ 2cosx -
sinx
= 1
17.
4 4 1
sin x cos x 2
+ =
18.
4 4
sin x cos x cos2x+ =
19.
4 4 x4 4
1
sin x sin
π
+
+ =
20.
2 2 2
2 2 3
sin x sin x sin x
3 3 2
π π
+ − + + =
21.
( )
6 6 4 4
5
sin x cos x sin x cos x
6
+ = +
22.
6 6 1
2
sin x cos x sinxcosx 0+ + =
23.
4 4 4 4
4sin x cos x sin x cos 4x+ = +
24.
( )
24 4 2
1
2sin x cos x sin xcos x sinxcosx+ = +
25.
3 3 2
cos xcos3x sin xsin3x= 4
+
25.
3 3 3
cos 4x cos xcos3x sin xsin3x=+
D¹ng 3: Ph ¬ng tr×nh bËc nhÊt ®èi víi sinx vµ cosx
1. NhËn d¹ng:
2
a.sinx b.cosx c+ =
Nguyen Tat Mao Sinh Vien dai Hoc Hang Hai Viet Nam
2. Ph ¬ng ph¸p:
§¨c biÖt :
1.
π π
sinx + 3cosx = 2sin(x + ) = 2cos(x - )
3 6
2.
sin cos 2 sin( ) 2 cos( )
4 4
x x x x
π π
± = ± =
m
3.
π π
sinx - 3cosx = 2sin(x - ) = -2cos(x + )
3 6
gi¶i ph¬ng tr×nh:
1.
3cosx sinx 2− =
, 2.
cosx 3sinx 1
− = −
3.
3
3sin3x 3cos9x 1 4sin 3x
− = +
, 4.
4 4
1
sin x cos (x )
4 4
π
+ + =
5.
3(1 cos 2 ) cos
2sin
−=
xx
x
, 6.
21
sin 2 sin 2
+ =x x
7.
1
3sinx + cosx = cosx
8.
tan 3cot 4(sin 3 cos )
− = +
x x x x
9.
cos7x - 3sin7x + 2 = 0
;
2π 6π
x ( ; )
5 7
∈
10. 2sin15x +
3
cos5x +
sin5x = 0 (4) 2.
6
11. sinx +3cosx + = 6
4sinx +3cosx +1
12.
1
3sinx + cosx = 3+ 3sinx + cosx +1
13. ( cos2x -
3
sin2x) -
3
sinx
– cosx + 4 = 0 14.
2
cosx - 2sinx.cosx = 3
2cos x + sinx -1
15.
2
1+ cosx + cos2x + cos3x 2
= (3- 3sinx)
2cos x + cosx -1 3
16.
cos7x sin5x 3(cos5x sin7x)
− = −
17. T×m GTLN vµ GTNN cña c¸c hµm sè sau:
3
C¸ch 1: asinx + bcosx = c
§Æt cosx=
2 2
a
a + b
; sinx=
2 2
b
a + b
2 2
a + b sin(x +α) = c
⇒
C¸ch 2:
b
a sinx + cosx = c
a
§Æt
b= tanα a sinx + cosx.tanα = c
a
⇒
c
sin(x +α) = cosα
a
⇔
C¸ch 3: §Æt
x
t = tan 2
ta cã
2
2 2
2t 1- t
sinx = ; cosx =
1+ t 1+ t
2
(b + c)t - 2at - b + c = 0
⇒
Chó ý: §iÒu kiÖn PT cã nghiÖm:
2 2 2
a + b c
≥
Nguyen Tat Mao Sinh Vien dai Hoc Hang Hai Viet Nam
a. y = 2sinx + 3cosx + 1 b.
1 cosx
ysinx cosx 2
−
=+ +
c.
2 cosx
ysinx cosx 2
+
=+ −
D¹ng 4: Ph ¬ng tr×nh ®¼ng cÊp ®èi víi sinx vµ cosx
1. NhËn d¹ng:
2. Ph ¬ng ph¸p:
Gi¶i ph¬ng tr×nh
1. 3sin2x -
3
sinxcosx+2cos2x cosx=2 2. 4 sin2x + 3
3
sinxcosx - 2cos2x=4
3. 3 sin2x+5 cos2x-2cos2x - 4sin2x=0 4. sinx - 4sin3x + cosx = 0
5. 2 sin2x + 6sinxcosx + 2(1 +
3
)cos2x – 5 -
3
= 0
6. (tanx - 1)(3tan2x + 2tanx + 1) =0 7. sin3x - sinx + cosx – sinx = 0
8. tanxsin2x - 2sin2x = 3(cos2x + sinxcosx) 9. 3cos4x - 4sin2xcos2x + sin4x = 0
10. 4cos3x + 2sin3x - 3sinx = 0 11. 2cos3x = sin3x
12. cos3x - sin3x = cosx + sinx 13. sinxsin2x + sin3x = 6cos3x
14. sin3(x -
π
/4) =
2
sinx
D¹ng 5: Ph ¬ng tr×nh ®èi xøng ®èi víi sinx vµ cosx
1. NhËn d¹ng:
2. Ph ¬ng ph¸p:
4
( )
( )
a sinx cosx b.sinxcosx c
a sinx cosx b.sinxcosx c
+ + =
− + =
2 2
3 2 2
a.sinx b.cosx 0 (1)
a.sin x b.sinxcosx c.cos x d (2)
a.sin x b.sin xcosx c.sinxcos x d.sinx e.cosx 0 (3)
+ =
+ + =
+ + + + =
§¼ng cÊp bËc 2: asin2x + bsinx.cosx + c cos2x = 0
C¸ch 1: Thö víi cosx = 0; víi cosx
≠
0, chia 2 vÕ cho cos2x ta ®-
îc:
atan2x + btanx + c = d(tan2x + 1)
C¸ch 2: ¸p dông c«ng thøc h¹ bËc
§¼ng cÊp bËc 3: asin3x + bcos3x + c(sinx + cosx) = 0
HoÆc asin3x + b.cos3x + csin2xcosx + dsinxcos2x = 0
* a(sin x + cosx) + bsinxcosx = c ®Æt t = sin x + cosx
t 2≤
⇒
at + b
2
t -1
2
= c
⇔
bt2 + 2at – 2c – b
= 0
* a(sin x - cosx) + bsinxcosx = c ®Æt t = sin x - cosx
t 2≤
⇒
at + b
2
1- t
2
= c
⇔
bt2 - 2at + 2c – b = 0
Nguyen Tat Mao Sinh Vien dai Hoc Hang Hai Viet Nam
1. 2(sinx +cosx) + sin2x + 1 = 0 2. sinxcosx = 6(sinx – cosx – 1)
3.
sin2x 2sin x 1
4
π
+ − =
3.
tanx 2 2sinx 1− =
1. 1 + tanx = 2sinx +
1
cos x
2. sin x + cosx=
1
tanx
-
1
cot x
3. sin3x + cos3x = 2sinxcosx + sin x + cosx 4. 1- sin3x+ cos3x = sin2x
5. 2sinx+cotx=2 sin2x+1 6.
2
sin2x(sin x + cosx) = 2
7. (1+sin x)(1+cosx)=2 8.
2
(sin x + cosx) = tanx + cotx
9. 1 + sin3 2x + cos32 x =
3
2
sin 4x 10.* 3(cotx - cosx) - 5(tanx - sin x) = 2
11.* cos4x + sin4x - 2(1 - sin2xcos2x)sinxcosx - (sinx + cosx) = 0
12.
sin cos 4sin 2 1x x x
− + =
13. sinxcosx +
sinx + cosx
= 1
14. cosx +
1
cosx
+ sinx +
1
sinx
=
10
3
D¹ng 6: Ph ¬ng tr×nh ®èi xøng ®èi víi sinx vµ cosx
Gi¶i ph¬ng tr×nh
1/ sin2 x + sin23x = cos22x + cos24x 2/ cos2x + cos22x + cos23x + cos24x = 3/2
3/ sin2x + sin23x - 3cos22x=0 4/ cos3x + sin7x = 2sin2(
π 5x
+
4 2
) - 2cos2
9
2
x
5/ cos4x – 5sin4x = 1 6/ 4sin3x - 1 = 3 -
3
cos3x
7/ sin22x + sin24x = sin26x 8/ sin2x = cos22x + cos23x
9/ (sin22x + cos42x - 1):
sinxcosx
= 0 10/ 2cos22x + cos2x = 4
sin22xcos2x
11/ sin3xcos3x +cos3xsin3x=sin34x 12/ 8cos3(x +
π
3
) =
cos3x
13/
sin5x
5sinx
= 1 14/ cos7x + sin22x =
cos22x - cosx 15/ sin2x + sin22x + sin23x =
3/2 16/ 3cos4x – 2cos23x =1
17/ sin24 x+ sin23x= cos22x+ cos2x víi
x (0;π)∈
18/ sin24x - cos26x = sin(
10,5π +10x
) víi
π
x (0; )
2
∈
19/ 4sin3xcos3x + 4cos3x sin3x + 3
3
cos4x = 3
20/ cos4xsinx - sin22x = 4sin2(
4 2
x
π
−
) -
7
2
víi
x -1
< 3
21/ 2cos32x - 4cos3xcos3x + cos6x - 4sin3xsin3x = 0
22/ cos10x + 2cos24x + 6cos3xcosx = cosx + 8cosxcos23x
5
C«ng thøc h¹ bËc 2 cos2x =
1 cos 2
2
x
+
; sin2x=
1- cos2x
2
C«ng thøc h¹ bËc 3 cos3x=
3cosx + cos3x
4
; sin3
x=
3sinx -sin3x
4
