PHÖÔNG TRÌNH LÖÔÏNG GIAÙC

H???ng da?n gia?i CDBT t?? ca?c ?TQG Toa?n ho?c ?

Chuye?n ?e? 2:

L???NG GIA?C

Va?n ?e? 1:

PH??NG TR?NH L???NG GIA?C

A. PH??NG PHA?P GIA?I

1. Ph??ng tr?nh l???ng gia?c c? ba?n

cosx = cos x = + k2

sinx = sin

x k2 x k2

tanx = tan x = + k

cotx = cot x = + k

(v??i k )

2. Ph??ng tr?nh ba?c hai ?o?i v??i mo?t ha?m so? l???ng gia?c asin2x + bsinx + c = 0. ?a?t t = sinx, t 1 acos2x + bcosx + c = 0. ?a?t t = cosx, t 1 atan2x + btanx + c = 0. ?a?t t = tanx acot2x + bcotx + c = 0. ?a?t t = cotx

3. Ph??ng tr?nh ba?c nha?t ?o?i v??i sinx, cosx

asinx + bcosx = c (*) ?ie?u kie?n co? nghie?m: a2 + b2 c2

Ca?ch 1: Chia hai ve? cho a2 b2 0

(*) a sinx + b cosx = c

a2 b2

a2 b2

a2 b2

a 2 b 2

Do

a2 b2

+

a2

b2

= 1

Ne?n co? the? ?a?t a = cos, b = sin

a2 b2

a2 b2

Khi ?o?:

(*) sinxcos + sincosx = c sin(x + ) = c

a2 b2

a2 b2

Ca?ch 2: Chia hai ve? cho a (gia? s?? a 0)

(*) sinx + b cosx = c

a

a

?a?t b = tan. Khi ?o?: (*) sinx + sin cosx = c

a

cos

a

70

TT Luyn Thi i Hc VNH VIN

sinx cos + sin cosx = c cos sin(x + ) = c cos

a

a

Ca?ch 3: ?a?t a?n so? phu?.

Xe?t x = (2k + 1) v??i (k ) co? la? nghie?m 0

Xe?t x (2k + 1) v??i (k )

?a?t t = tan x 2

Khi

?o?:

(*)

a

1

2t t

2

+

b

1 1

t2 t2

= c (b + c)t2 ? 2at + c ? b = 0

4. Ph??ng tr?nh ?o?i x??ng: a(sinx + cosx) + bsinxcosx + c = 0

?a?t t = sinx + cosx =

2

cos

x

4

?ie?u kie?n t 2

Khi ?o?: t2 = 1 + 2sinxcosx sinxcosx = t2 1 2

Thay va?o ph??ng tr?nh ta ????c ph??ng tr?nh ?a?i so? theo t. Chu? y?: a(sinx cosx) + bsinxcosx + c = 0

?a?t t = sinx ? cosx (v??i t 2 )

5. Ph??ng tr?nh ?a?ng ca?p ba?c 2 ?o?i v??i sinx, cosx asin2x + bsinxcosx + ccos2x = 0

Xe?t cosx = 0 x = + k (k ) co? la? nghie?m kho?ng? 2

Xe?t cosx 0. Chia 2 ve? cho cos2x ta thu ????c ph??ng tr?nh ba?c 2 theo tanx.

Chu? y?: Ne?u la? ph??ng tr?nh ?a?ng ca?p ba?c k ?o?i v??i sinx, cosx th? ta xe?t cosx = 0 va? xe?t cosx 0 chia 2 ve? cu?a ph??ng tr?nh cho coskx va? ta thu ????c mo?t ph??ng tr?nh ba?c k theo tanx.

B. ?E? THI

Ba?i 1: ?A?I HO?C KHO?I A NA?M 2011

Gia?i

ph??ng

tr?nh:

1 sin 2x cos2x 1 cot2 x

2 sin x.sin 2x .

Gia?i

?ie?u kie?n: sinx 0. Khi ?o?:

(1)

1

sin 2x 1

cos 2x

2 sin x. 2sin x cosx

sin2 x

71

H???ng da?n gia?i CDBT t?? ca?c ?TQG Toa?n ho?c ?

sin2 x 1 sin2x cos2x 2 2 sin2 x.cosx

1 sin2x cos2x 2 2 cosx (v? sinx 0)

2cos2 x 2sin x cosx 2 2 cosx 0

cosx 0 cosx sin x 2

cos x

0

sin

x

4

1

x k x k2 (k Z) (Tho?a ?ie?u kie?n sinx 0).

2

4

Va?y nghie?m cu?a (1) la? x k x k2 (k Z).

2

4

Ba?i 2: ?A?I HO?C KHO?I B NA?M 2011

Gia?i ph??ng tr?nh: sin2xcosx sinxcosx cos2x sinx cosx

Gia?i sin2xcosx sinxcosx cos2x sinx cosx 2sinx.cos2x + sinx.cosx = 2cos2x ? 1 + sinx + cosx sinx.cosx(2cosx + 1) = cosx(2cosx + 1) + sinx ? 1 cosx (2cosx + 1)(sinx ? 1) = sinx ? 1 sinx ? 1 = 0 hoa?c cosx (2cosx + 1) = 1 sinx = 1 hoa?c 2cos2x + cosx ? 1 = 0

sinx = 1 hoa?c cosx = ?1 hoa?c cosx = 1 2

x k2 hoa?c x k2 hoa?c x k2

2

3

x k2 hoa?c x k 2 (k Z)

2

33

Ba?i 3: ?A?I HO?C KHO?I D NA?M 2011

Gia?i ph??ng tr?nh: sin 2x 2 cosx sin x 1 0 tan x 3

Gia?i

sin 2x 2 cosx sin x 1 0 . ?ie?u kie?n: tanx 3 va? cosx 0. tan x 3

sin2x 2cosx sinx 1 0 2sin x cosx 2cosx sin x 1 0

72

TT Luyn Thi i Hc VNH VIN

2cosxsinx 1 sinx 1 0 sinx 12cosx 1 0

sin x 1 (Loa?i v? khi ?o? cosx = 0)

cos

x

1 2

x k2 (k Z). 3

So v??i ?ie?u kie?n ta ????c nghie?m cu?a ph??ng tr?nh la? x k2 (k Z). 3

Ba?i 4: CAO ?A?NG KHO?I A, B, D NA?M 2011 Gia?i ph??ng tr?nh: cos4x + 12sin2x ? 1 = 0.

Gia?i cos4x + 12sin2x ? 1 = 0 2cos22x ? 1 + 6(1 ? cos2x) ? 1 = 0 cos22x ? 3cos2x + 2 = 0 cos2x = 1 hay cos2x = 2 (loi) 2x = k2 x = k (k Z). Ba?i 5: ?A?I HO?C KHO?I A NA?M 2010

Gia?i ph??ng tr?nh:

(1

sin

x

cos

2x)

sin

x

4

1

cos x

1 tan x

2

Gia?i

?ie?u kie?n: cosx 0 va? tanx ? 1

V??i ?ie?u kie?n tre?n, ph??ng tr?nh ?a? cho t??ng ???ng:

(1 sin x cos2x).(sin x cosx) cosx 1 tan x

(1 sin x cos2x).(sin x cosx) cosx cosx sin x cosx

1 sin x cos2x 1 sin x cos2x 0

2sin2 x sin x 1 0 sin x 1(loa?i) hay sin x 1 2

x k2 hay x 7 k2 (k Z)

6

6

Ba?i 6: ?A?I HO?C KHO?I B NA?M 2010

Gia?i ph??ng tr?nh (sin 2x + cos 2x) cosx + 2cos2x ? sin x = 0

Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng:

(2sinxcosx + cos2x)cosx + 2cos2x ? sinx = 0 cos2x (cosx + 2) + sinx (2cos2x ? 1) = 0

cos2x (cosx + 2) + sinx.cos2x = 0

73

H???ng da?n gia?i CDBT t?? ca?c ?TQG Toa?n ho?c ?

cos2x (cosx + sinx + 2) = 0

cos2x 0 cosx sin x 2 0

(vn)

2x = k (k ) x = k (k ) .

2

42

Ba?i 7: ?A?I HO?C KHO?I D NA?M 2010

Gia?i ph??ng tr?nh sin2x cos2x 3sinx cosx 1 0 Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng: 2sin x cos x 1 2sin2 x 3sin x cos x 1 0

cos x(2sin x 1) 2sin2 x 3sin x 2 0

cos x(2sin x 1) (2sin x 1)(sin x 2) 0

(2sin x 1)(cos x sin x 2) 0

sin

x

1 2

cos x sin x 2 (VN)

x x

k2 6 5 k2 6

(k

).

Ba?i 8: CAO ?A?NG KHO?I A, B, D NA?M 2010

Gia?i ph??ng tr?nh 4 cos 5x cos 3x 2(8sin x 1)cosx 5 . 22

Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng: 2(cos4x cosx) 16sin x cosx 2cosx 5

2cos4x 8sin2x 5 2 4sin2 2x 8sin2x 5

4sin22x ? 8sin2x + 3 = 0 sin 2x 3 (loa?i ) hay sin 2x 1

2

2

2x k2 hay 2x 5 k2

6

6

x k hay x 5 k (k ) .

12

12

Ba?i 9: ?A?I HO?C KHO?I A NA?M 2009

Gia?i

ph??ng

tr?nh:

1 2sin x cosx 1 2sin x1 sin x

3.

74

TT Luyn Thi i Hc VNH VIN

Gia?i ?ie?u kie?n: sinx 1 va? sinx 1 (*)

2 V??i ?ie?u kie?n tre?n, ph??ng tr?nh ?a? cho t??ng ???ng:

(1 ? 2sinx)cosx = 3 1 2sin x1 sin x

cosx 3 sin x sin2x 3 cos2x

cos

x

3

cos

2x

6

x k2 hoa?c x k 2 (k )

2

18 3

Ke?t h??p (*), ta ????c nghie?m: x k 2 k

18 3

Ba?i 10: ?A?I HO?C KHO?I B NA?M 2009

Gia?i ph??ng tr?nh: sinx + cosxsin2x + 3 cos3x 2 cos4x sin3 x

Gia?i Ph??ng tr?nh ?a? cho t??ng ???ng:

(1 ? 2sin2x)sinx + cosxsin2x + 3 cos3x 2cos4x

sinxcos2x + cosxsin2x + 3 cos3x 2cos4x

sin3x +

3

cos

3x

2

cos

4x

cos

3x

6

cos

4x

4x = 3x k2 hoa?c 4x 3x k2 (k )

6

6

Va?y:

x

=

6

k2;

x

42

k

2 7

k

.

Ba?i 11: ?A?I HO?C KHO?I D NA?M 2009

Gia?i ph??ng tr?nh: 3 cos5x 2sin3x cos2x sin x 0 Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng:

3 cos5x sin5x sin x sin x 0

3 2

cos5x 1 sin 5x sin x 2

sin

3

5x

sin x

5x x k2 hay 5x x k2 (k )

3

3

Va?y: x = k hay x k k

18 3

62

75

H???ng da?n gia?i CDBT t?? ca?c ?TQG Toa?n ho?c ?

Ba?i 12: CAO ?A?NG KHO?I A, B, D NA?M 2009 Gia?i ph??ng tr?nh (1 + 2sinx)2cosx = 1 + sinx + cosx

Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng:

(1 + 4sinx + 4sin2x)cosx = 1 + sinx + cosx

cosx + 4sinxcosx + 4sin2xcosx = 1 + sinx + cosx

1 + sinx = 0 hay 4sinxcosx = 1

sinx = 1 hay sin2x = 1 2

x k2 hay x k hay x 5 k (v??i k ) .

2

12

12

Ba?i 13: ?A?I HO?C KHO?I A NA?M 2008

Gia?i ph??ng tr?nh:

1 sin x

sin

1 x

3 2

4

sin

7 4

x

Gia?i

Ta

co?:

sin

x

3 2

cos x

?ie?u kie?n:

sin x 0 cos x 0

sin2x 0

V??i ?ie?u kie?n tre?n, ph??ng tr?nh ?a? cho t??ng ???ng:

1 sin

x

1 cos

x

4

sin

x

4

cosx sin x 2 2 sin x cosxsin x cosx

cosx sin x 1 2 sin2x 0

cos x sin x 0 tan x 1

x

4

k

sin 2x

1

2

sin

2x

2

x

8

k

2

x

5

k

(k

) .

8

Ba?i 14: ?A?I HO?C KHO?I B NA?M 2008

Gia?i ph??ng tr?nh: sin3 x 3 cos3 x sin x cos2 x 3 sin2 x cosx

Gia?i

sin3 x 3 cos3 x sin x.cos2 x 3 sin2 x.cosx (1)

76

TT Luyn Thi i Hc VNH VIN

Ca?ch 1: Ph??ng tr?nh ?a? cho t??ng ???ng: sin x(cos2 x sin2 x) 3 cosx(cos2 x sin2 x) 0

cos2 x sin2 x sin x 3 cosx 0

cos2x 0 tan x 3

x

4

k 2

x

3

k

(k )

Nghie?m cu?a ph??ng tr?nh la?: x k va? x k

42

3

(k )

Ca?ch 2:

cosx = 0 kho?ng pha?i la? nghie?m cu?a ph??ng tr?nh (1). Chia hai ve? cu?a ph??ng tr?nh (1) cho cos3x ta ????c:

tan3 x 3 tan x 3 tan3 x

(tan x

3

)(tan2

x

1)

0

tan tan

x x

1

3

x x

3 4

k k

k

Ba?i 15: ?A?I HO?C KHO?I D NA?M 2008

Gia?i ph??ng tr?nh: 2sinx(1 + cos2x) + sin2x = 1 + 2cosx.

Gia?i

Ph??ng tr?nh ?a? cho t??ng ???ng: 4sinx.cos2x + sin2x ? 1 ? 2cosx = 0

2cosx(2sinxcosx ? 1) + (sin2x ? 1) = 0

(sin2x ? 1)(2cosx + 1) = 0

sin 2x 1hay cosx 1 x k hayx 2 k2 hay x 2 k2 (k )

2

4

3

3

Ba?i 16: CAO ?A?NG KHO?I A, B, D NA?M 2008

Gia?i ph??ng tr?nh: sin3x 3 cos3x 2sin2x .

Gia?i Ph??ng tr?nh ?a? cho t??ng ???ng:

1 sin3x 3 cos3x sin 2x cos sin3x sin cos3x sin 2x

2

2

3

3

sin

3x

3

sin

2x

77

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download