
Chöông 2: PHÖÔNG TRÌNH LÖÔÏNG GIAÙC CÔ BAÛN
=+ π
⎡
=⇔
⎢=π− + π
⎣
uvk2
sin u sin v uvk2
cos u cos v u v k2=⇔=±+π
π
⎧≠+π
⎪
=⇔
⎨
⎪=+ π
⎩
uk
tgu tgv 2
uvk'
(
)
k,k ' Z∈
uk
cot gu cot gv uvk'
≠π
⎧
=⇔
⎨=+ π
⎩
Ñaëc bieät : si
n u 0 u k=⇔=π π
=
⇔=+πco
s u 0 u k
2
(
sin u 1 u k2 k Z
2
π
=⇔= + π ∈
)
cos u 1 u k2
=
⇔= π
()
kZ∈
sin u 1 u k2
2
π
=− ⇔ =− + π
cos u 1 u k2
=
−⇔ =π+ π
Chuù yù :
sin u 0 cos u 1≠⇔ ≠±
cos u 0 sin u 1≠⇔ ≠±
Baøi 28 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái D, naêm 2002)
[
]
x0,14∈ nghieäm ñuùng phöông trình Tìm
(
)
cos 3x 4 cos 2x 3 cos x 4 0 *−+−=
Ta coù (*) : ⇔
()
(
)
32
4 cos x 3 cos x 4 2 cos x 1 3 cos x 4 0
−
−−+−=
⇔ 32
4cos x 8cos x 0
−
= ⇔
(
)
2
4cos x cosx 2 0
−
=
⇔
(
)
==cosx 0hay cosx 2 loaïi vìcosx 1≤
⇔
()
xkk
2
π
=+π∈Z
Ta coù :
[]
x0,14 0 k 1
24
π
∈⇔≤+π≤
⇔ k14
22
ππ
−≤π≤ − ⇔ 1141
0, 5 k 3, 9
22
−=−≤≤−≈
π
Maø k neân Z∈
{
}
k. Do ñoù :
0,1,2,3∈357
x ,,,
2222
π
πππ
⎧
⎫
∈
⎨
⎬
⎩⎭
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 2 cos x 1−
⇔
()( )
2cos x 1 2sin x cos x sin x 0
−
+−
⎡⎤
⎣⎦
=
)
⇔
()(
2cosx 1 sinx cosx 0
−
+=
⇔ 1
cos x sin x cos x
2
=∨ =−
⇔ cos x cos tgx 1 tg
34
ππ
⎛⎞
=∨=−=−
⎜⎟
⎝⎠
⇔
()
ππ
=± + π∨ =− + π ∈xk2xk,k
34
Z
Baøi 30 : Giaûi phöông trình
+
++=cos x cos 2x cos 3x cos 4x 0(*)
Ta coù (*)
⇔
()
(
)
cos x cos 4x cos 2x cos 3x 0+++=
⇔ 5x 3x 5x x
2cos .cos 2cos .cos 0
22 22
+=
⇔ 5x 3x x
2cos cos cos 0
22 2
⎛⎞
+=
⎜⎟
⎝⎠
⇔ 5x x
4 cos cos x cos 0
22
=
⇔ 5x x
cos 0 cos x 0 cos 0
22
=
∨=∨ =
⇔ ππ π
=+π∨=+π∨=+π
5x x
kx k k
22 2 22
⇔
()
ππ π
=+ ∨=+π∨=π+π ∈
2k
xxkx2,
55 2 kZ
Baøi 31: Giaûi phöông trình
(
)
22 2 2
sin x sin 3x cos 2x cos 4x *+=+
Ta coù (*) ⇔
()()()()
1111
1 cos 2x 1 cos 6x 1 cos 4x 1 cos 8x
2222
−+−=+++
⇔
()
cos2x cos6x cos4x cos8x−+ =+
⇔
2 cos 4x cos 2x 2 cos 6x cos 2x−=
⇔
(
)
2cos2x cos6x cos 4x 0+=
⇔ 4 cos 2x cos5x cos x 0
=
⇔ cos 2x 0 cos5x 0 cos x 0
=
∨=∨=
⇔ ππ π
=+π∨ +π∨=+π∈
2x k 5x k x k , k
22 2
⇔ ππ π π π
=+ ∨= + ∨=+πk
kk
xx x
42 105 2
∈
,k
Baøi 32 : Cho phöông trình
()
π
⎛⎞
−= −−
⎜⎟
⎝⎠
22
x7
sin x.cos 4x sin 2x 4 sin *
42 2
Tìm caùc nghieäm cuûa phöông trình thoûa:
−
<x1 3

Ta coù : (*)⇔
()
17
sin x.cos 4x 1 cos4x 2 1 cos x
22
⎡π⎤
⎛⎞
2
−
−=−−
⎜⎟
⎢⎥
⎝⎠
⎣⎦
−
⇔ −+ =−−
11 3
sin x cos 4x cos 4x 2 sin x
22 2
⇔ 1
sin x cos 4x cos 4x 1 2sin x 0
2
+++=
⇔ ⎛⎞⎛⎞
++ +=
⎜⎟⎜⎟
⎝⎠⎝⎠
11
cos 4x sin x 2 sin x 0
22
⇔
()
1
cos 4x 2 sin x 0
2
⎛⎞
+
+=
⎜⎟
⎝⎠
⇔
()
cos 4x 2 loaïi
1
sin x sin
26
=−⎡
⎢
π
⎛
⎢=− = −
⎜⎟
⎢⎝⎠
⎣
⎞
⇔
π
⎡
=
−+ π
⎢
⎢π
⎢
=
+π
⎢
⎣
xk
6
7
x2
6
2
h
Ta coù :
−
<x1 3
⇔ ⇔
3x13−< − < 2x4
−
<<
Vaäy : 2k2
6
π
−<− + π<4
⇔ 22k 4
66
ππ
−< π<+ ⇔ 11 21
k
12 12
−<<+
ππ
Do k neân k = 0. Vaäy Z∈x6
π
=
−
π
−< + π<
7
2h2
64
⇔
π
π
−− < π< − ⇔− − < < −
π
π
77172
2h24 h
6612
7
12
⇒h = 0 ⇒π
=7
x6.Toùm laïi
−
ππ
==
7
xhayx
66
Caùch khaùc :
−
π
=− ⇔ = − + π ∈
k
1
sin x x ( 1) k , k
26
Vaäy : −π − −
−<− +π< ⇔ <− + <
π
π
kk
21
2(1) k 4 (1) k
66
4
⇔ k=0 vaø k = 1. Töông öùng vôùi
−
ππ
==
7
xhayx
66
Baøi 33 : Giaûi phöông trình
(
)
33 3
sin x cos 3x cos x sin 3x sin 4x *+=
Ta coù : (*)⇔
()
(
)
33 3 3 3
sin x 4 cos x 3cos x cos x 3sin x 4 sin x sin 4x−+ − =
⇔
33 3 3 33 3
4 sin x cos x 3sin x cos x 3sin x cos x 4 sin x cos x sin 4x−+− =
⇔
()
22 3
3sin x cos x cos x sin x sin 4x−=
⇔ 3
3sin 2x cos 2x sin 4x
2=

⇔ 3
3sin 4x sin 4x
4=
⇔ 3
3sin 4x 4 sin 4x 0
−
=
⇔ sin12x = 0
⇔ ⇔
12x k=π
()
k
xk
12 Z
π
=∈
Baøi 34 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái B, naêm 2002)
Giaûi phöông trình :
(
)
22 22
sin 3x cos 4x sin 5x cos 6a *−=−
Ta coù : (*)⇔
()()()()
11 1 1
1 cos 6x 1 cos 8x 1 cos10x 1 cos12x
22 2 2
−−+=− −+
⇔
cos 6x cos 8x cos10x cos12x+= +
⇔
2cos7xcosx 2cos11xcosx=
⇔
(
)
2cos x cos7x cos11x 0−=
⇔
cos x 0 cos7x cos11x=∨ =
⇔ π
=+π∨ =± + πxk7x11xk
22
⇔ πππ
=+π∨=− ∨= ∈
kk
xkx x,k
229
Baøi 35 : Giaûi phöông trình
()()
sin x sin 3x sin 2x cos x cos 3x cos 2x++=++
⇔
2sin 2x cos x sin 2x 2cos 2x cos x cos 2x+= +
⇔
()
(
)
+= +sin 2x 2 cos x 1 cos 2x 2 cos x 1
⇔
()( )
2 cos x 1 sin 2x cos 2x 0
+
−=
⇔ 12
cos x cos sin 2x cos 2x
23
π
=− = ∨ =
⇔ 2
xk2tg2x1
34
tg
π
π
=± + π∨ = =
⇔
()
π
ππ
=± + π∨ = + ∈
2
xk2xk,k
382
Z
Baøi 36: Giaûi phöông trình
(
)
++ =+
23
cos10x 2 cos 4x 6 cos 3x.cos x cos x 8 cos x.cos 3x *
Ta coù : (*)⇔
(
)
(
)
3
cos10x 1 cos8x cos x 2cos x 4cos 3x 3cos 3x++ = + −
⇔
()
cos10x cos 8x 1 cos x 2 cos x.cos 9x++=+
⇔
2 cos 9x cos x 1 cos x 2 cos x.cos 9x+= +
⇔
cos x 1=
⇔
(
)
xk2kZ=π∈
Baøi 37 : Giaûi phöông trình

(
)
33 2
4 sin x 3cos x 3sin x sin x cos x 0 *+−− =
Ta coù : (*) ⇔
()
(
)
22
sin x 4 sin x 3 cos x sin x 3cos x 0
2
−
−−=
⇔
()
(
)
⎡⎤
−− − − =
⎣⎦
22
sin x 4 sin x 3 cos x sin x 3 1 sin x 0
2
=
=
⇔
()
()
2
4sin x 3 sinx cosx 0−−
⇔
()( )
2 1 cos 2x 3 sin x cos x 0−− −
⎡⎤
⎣⎦
⇔
12
cos 2x cos
23
sin x cos x
π
⎡=− =
⎢
⎢=
⎣
⇔
2
2x k2
3
tgx 1
π
⎡=± + π
⎢
⎢=
⎣
⇔
xk
3
xk
4
π
⎡
=
±+π
⎢
⎢π
⎢
=
+π
⎢
⎣
(
)
kZ∈
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 : (*) ⇔
2
sin x cos x 2sin x cos x 2cos x 0++ + =
⇔
(
)
sin x cos x 2 cos x sin x cos x 0++ + =
⇔
()(
sin x cos x 1 2 cos x 0++
)
=
⇔
sin x cos x
12
cos 2x cos
23
=−
⎡
⎢
π
⎢=− =
⎣
⇔
tgx 1
2
xk
3
=−
⎡
⎢π
⎢=± + π
⎣2
⇔
xk
4
2
xk2
3
π
⎡=− + π
⎢
⎢π
⎢=± + π
⎢
⎣
()
kZ∈
Baøi 39 : Giaûi phöông trình
()( )
(
)
2
2sinx 1 3cos4x 2sinx 4 4cos x 3 *++−+=
Ta coù : (*) ⇔
()
(
)
(
)
2
2sinx 1 3cos4x 2sinx 4 4 1 sin x 3 0++−+−−=
⇔
()( )
(
)
(
)
2sinx 1 3cos4x 2sinx 4 1 2sinx 1 2sinx 0
+
+−++ − =
⇔
()
(
)
2sinx 1 3cos4x 2sinx 4 1 2sinx 0
+
+−+−
⎡⎤
⎣⎦
=
=
⇔
()()
3cos4x 1 2sinx 1 0−+
⇔ 1
cos 4x 1 sin x sin
26
π
⎛⎞
=∨ =− = −
⎜⎟
⎝⎠

