Chuyên Đề 1: PT bậc 2 với một HSLG
1) 2sin2x - cos2x - 4sinx + 2 = 0
2) 9cos2x - 5sin2x - 5cosx + 4 = 0
3) 5sinx(sinx - 1) - cos2x = 3
4) cos2(3x +
2
π) – cos23x – 3cos(
2
π - 3x) + 2 = 0 5)[cđsphn_97] cos2x + sin2x + 2cosx + 1 = 0
6) 3cos2x + 2(1 + 2 + sinx)sinx – (3 + 2) = 0
7) tg2x + ( 3 - 1)tgx – 3 = 0
sin
3
2 = gx+
x
2 cos 2 cot
4 sin 2 cot 3 2 cos
=
−
+ +
x x
g
x x
g x
10)[ĐHBKHN_94]
0 cos
2 cos 3 9 sin
6
2
sin
=
−
− +
x
x x
x
11)[ĐH Thuỷ Sản Nha Trang_01]
1 1
2 sin
) 2 (sin
sin 3 ) sin 2
(cos
−
+ +
+
x
x x x
x
x
Chuyên Đề 2: PT bậc 3 với một HSLG
1) 4sin3x – 8sin2x + sinx + 3 = 0
2)[ĐH Luật HN_00] 4(sin3x – cos2x) = 5(sinx – 1)
3)4 cos 3x+ ( 6 − 2 3 ) cos 2x− ( 4 + 3 3 ) cosx+ 2 3 = 0
4) cos3x + 3cos2x = 2(1 + cosx)
5) 2tg3x + 5tg2x – 23tgx + 10 = 0
6) 6tg3x+ ( 3 − 2 3 )tg2x− ( 3 + 3 )tgx+ 3 = 0
7) tg3x – tgx = 2
8) cotg3x + 2cotg2x – 3cotgx - 6 = 0
9) 2cotg3x – cotg2x – 13cotgx – 6 = 0
10)[ĐHNN HN_00] 2cos2x – 8cosx + 7 = 1/cosx
Chuyên Đề 3: PT bậc nhất đối với Sin và cos
1) 3 cos 3x+ sin 3x= 2
2)[ĐH Mỏ_95] 3 sin 3x− 3 cos 9x= 1 + 4 sin 3 3x
3)[ĐH Mỹ Thuật Công Nghiệp HN_96]
x x x
x
xcos 5 3 sin 2 1 sin 7 sin 5
7
4)[ĐHKT_97] Tìm các nghiệm x )
7
6
; 5
2 ( π π
cos 7x− 3 sin 7x= − 2
5)[ĐHGT_00] 2 2 (sinx+ cosx) cosx= 3 + cos 2x
6 5 sin(
5 ) 6 sin(
4 ) 3 sin(
3 x−π + x+π + x+π =
7)[HVCNBCVT_01]
3 4 cos 3 3 3 sin cos 4 3 cos
sin
8) 2sin4x + 3cos2x + 16sin3xcosx – 5 = 0
9)[CĐHQ TPHCM_98] 4sin3x – 1= 3sinx - 3cos3x
10)[ĐH Kỹ Thuật Công Nghệ TPHCM_00]
cos2x - 3sin2x = 1 + sin2x
11)[ĐH Văn Lang TPHCM_98]
4(sin4x + cos4x) + 3sin4x = 2
12)[ĐHNN I_95]
x x
x
x 3 sin 2 sin 3 cos 2
cos
13)[ĐHTM_00] 3 sin 2x− 2 cos 2 x= 2 2 + 2 cos 2x
14)[ĐHSP Quy Nhơn_98]
2 cos 3 sin cos
3
Chuyên Đề 4: PT đcấp bậc 2 đối với sin và cos
1) sin2x + 2sinxcosx + 3cos2x - 3 = 0 2) sin2x – 3sinxcosx + 1 = 0
3) 4 3sinxcosx + 4cos2x = 2sin2x + 5/2
2 cos(
) 2
5 sin(
2 ) 3 ( sin
3 2 π−x + π +x π +x
) 0
2
3 ( sin
5)[ĐHAN_98]
a
x x
x
cos
1 cos
sin
b
x x
x
cos
1 cos
6 sin
6) cos2x – 3sinxcosx – 2sin2x – 1 = 0 7) 6sin2x + sinxcosx – cos2x = 2
Chuyên Đề 5: PT đcấp bậc 3 đối với sin và cos
1)[ĐHL_96] 4sin3x + 3cos3x – 3sinx – sin2xcosx = 0
2)[ĐHNT_96] cos3x – 4sin3x – 3cosxsin2x + sinx = 0 3)[ĐH Huế_98] cos3x + sinx – 3sin2xcosx = 0
4)[ĐH Đà Nẵng_99] cos3x – sin3x = sinx – cosx 5)[CĐSPTW1_01] 4cos3x + 2sin3x – 3sinx = 0 6)[HVKTQS_96] 2cos3x = sin3x
7)[ĐHD TPHCM_97] sinxsin2x + sin3x = 6cos3x 8)[ĐHY HN_99] sinx + cosx - 4sin3x = 0
9)[ĐHQGHN_96] 1 + 3sin2x = 2tgx 10)[ĐHNN B_99] sin2x(tgx + 1) = 3sinx(cosx – sinx) +3
11)[PVBCTT_98] x ) 2 sinx
4 ( sin
2 3 +π = 12)[ĐHQGHN_98] x ) cos 3x
3 ( cos
8 3 +π = 13)[ĐHQG TPHCM_98] x ) 2 sinx
4 ( sin 3 −π = 14)[ĐHYHN_95]
x
x x x
x
2 cos 2
cos 4 sin 5 cos 2 sin
Chuyên Đề 6: PT đối xứng và nửa đối xứng Với sin và cos
1) 2(sinx + cosx) - sinxcosx = 1 2) (1 – sinxcosx)(sinx + cosx) =
2 2
3)
3
10 sin
1 sin cos
1
x
x x
x
4) sin3x + cos3x =
2 2
5) 1 + sin3x + cos3x =
2
3
sin2x 6)[HVCTQG TPHCM] 2sin2x – 2(sinx + cosx) +1 = 0 7)[ĐH Huế D_00] sinxcosx + 2sinx + 2cosx = 2 8)[ĐHM_99] 1 + tgx = 2 2sinx
Trang 29) sinx + cosx = 1 sinx cos x
3
3 2
+ 10) sinx – cosx + 7sin2x = 1
11) ( 1 + 2 )(sinx− cosx) + 2 sinxcosx= 1 + 2
4 sin(
2 2 sin x+ x−π = 13) sinx− cosx + 4 sin 2x= 1
Chuyên Đề 7: PTLG đối xứng với tg và cotg
1) 3(tgx + cotgx) = 4
2)[ĐHCĐ_97] 2(sinx + cosx) = tgx + cotgx
3)[ĐHNN_97] cotgx – tgx = sinx + cosx
4)[ĐH Cần Thơ_D99] 3(tgx + cotgx) = 2(2 + sin2x)
5)[ĐHGT_95] tg2x + cotgx = 8cos2x
6)[ĐHQG_B96] tgx = cotgx + 2cotg32x
7)[ĐH Đông Đô_97] tgx + cotgx = 2(sin2x + cos2x)
8)[ĐH Đông Đô_99] cotgx = tgx + 2tg2x
9)[97II] 6tgx + 5cotg3x tg2x
10)[ĐHYHN_98] 2(cotg2x – cotg3x) = tg2x + cotg3x
11)[ĐHQG TPHCM_A96] tg2x – tgxtg3x = 2
12)[ĐHTH_A93] 3tg2x – 4tg3x = tg23xtg2x
13)[CĐ Hải Quan_00]
3tg2x + 4tgx + 4cotgx + 3cotg2x +2 = 0
14) tgx + tg2x + tg3x + cotgx + cotg2x + cotg3x = 6
15) tg2x – tg3x – tg5x = tg2xtg3xtg5x
16) tg22xtg23xtg5x = tg22x – tg23x + tg5x
17)[ĐHDHN_01]
tg2x.cotg22x.cotg3x = tg2x – cotg22x + cotg3x
18)[CĐGT_01]
tg2x.tg23x.tg4x = tg2 – tg23x + tg4x
19)[ĐHNT TPHCM_97] 2tgx + cotgx =
x
sin
2
3 + 20)[71 III] 3tg3x + cotg2x = 2tgx +
x
4 sin
2
21)[ĐHQG_A98] 2tgx +cotg2x = 2sin2x +
x
2 sin 1
22) 3tg6x -
x
8 sin
2
= 2tg2x – cotg4x 23)[130 III] cotg2x + cotg3x +
x x
xsin 2 sin 3 sin
1
= 0 24)[ĐHBK_98]
1 cot
) sin (cos 2 2
cot
1
−
−
=
x x x
g tgx
cos
1 sin
1
(
x
2
1 2
sin
cos sin 4 4
gx tgx
x
x x
+
= +
Chuyên Đề 8: PTLG Đxứng đối với sin 2n và
cos 2n
1)[ĐHBKHN_96] sin4x + cos4x = cos2x
2)[ĐH Huế_99] sin6x + cos6x = 7/16
3) sin6x + cos6x = sin 2x
4
4) sin6x + cos6x = cos4x
5)[HVCTQG TPHCM_00]
16(sin6x + cos6x – 1) + 3sin6x = 0 6)[ĐHQG_98] cos6x – sin6x =
8
13
cos22x 7)[ĐHKT_95]
8
5 ) 3 ( cos ) 3 ( sin 4 x + 4 x = 8)[ĐHCĐ_01] x x) 1 2 sinx
2 ( cos ) 2 (
9)
8
3 3 sin
1 3
cos
1
2
x x
Chuyên Đề 9: sử dụng ct hạ bậc
1) cos2x + cos22x + cos23x = 3/2 2) cos2x + cos22x + cos23x = 1 3)[ĐH Huế] sin2x + sin22x + sin23x = 3/2 4)[ĐHY_98] sin23x – sin22x – sin2x = 0 5)[ĐHQG_98] sin2x = cos22x + cos23x 6)[ĐH_B02] sin23x – cos24x = sin25x – cos26x 7) cos2x + cos22x + cos23x + cos24x = 2
8) cos2x + cos22x + cos23x + cos24x = 3/2 9)[Đ52II] cos2x = cos
3
4x
10)[Đ15III]
5
4 cos 3 5
3 cos 2
11)[Đ48II] sin22x – cos28x = 10 )
2
17 sin( π+ x
12)[ĐHD_99] sin24x – cos26x = sin( 10 , 5π+ 10x)
13)[ĐHTDTT_01]
cos3x+sin7x=
2
9 cos 2 ) 2
5 4 ( sin
− +
π
14)[ĐHHH_95] sin4x +
4
1 ) 4 ( cos 4 x+π = 15)[ĐHBKHN_95]
2sin2x(4sin4x – 1) = cos2x(7cos22x + 3cos2x – 4) 16)[ĐHXD_97]
x x
tg x tg
x
) 4 ( ) 4 (
2 cos 2
+
−
+
π π
11)[ĐHGT_99]
6 ( cot ) 3 ( cot 8
7
x g x
12)[ĐHGT_01]
Sin4x +
8
9 ) 4 ( sin ) 4 ( sin 4 x+π + 4 x−π = 13)[ĐHNT TPHCM_95] sin8x + cos8x=
16
17
cos22x 14)[HVKTMM_99] sin8x + cos8x =
32 17
15)[Vô Địch New York_73]
sin8x + cos8x =
128 97
16)[HVQY_97] sin82x + cos82x = 1/8 17)[ĐHKT_99] sin2x + sin23x = cos22x + cos24x 18)[Đ135II] cos3xcos3x + sin3xsin3x =
4 2
Trang 319)[Đ142III] cos3xcos3x + sin3xsin3x = cos34x
20)[ĐHNT_99] cos3xsin3x + sin3xcos3x = sin34x
21)[HVBCVT_01]
4sin3xsin3x + 4sin3xcos3x + 3 3cos4x = 3
22)[ĐHSP TPHCM_00]
2cos2x + 2cos22x + 2cos23x – 3 = cos4x(2sin2x +
1)
23)[ĐHN_01]
cos3xcos3x – sin3xsin3x = cos34x + 1/2
Chuyên Đề 10: sử dụng CT góc nhân đôi
1)[ĐHY_97] cos4x + sin6x = cos2x
2)[ĐHNN_97] cos2x + 5sinx + 2 = 0
3)[ĐHNN_99] 2sin3x – cos2x + cosx = 0
4)[Đ68II] 2cos3x + cos2x + sinx = 0
5)[Đ72II] cos4x – cos2x + 2sin6x = 0
6)[ĐHNT_95] 4cosx – 2cos2x – cos4x = 1
7)[ĐHHH_99] cos2x + 5 = 2(2 – cosx)(sinx- cosx)
8)[ĐHYHN_00] sin3x + cos3x = cos2x
9)[ĐH Huế_98] 2sin3x + cos2x = sinx
10)[ĐHQGHN_95] 4sin2x – 3cos2x = 3(4sinx – 1)
11)[Đ16III] Tìm nghiệm ; 3 )
2 (π π
∈
x x
2
7 cos(
3 ) 2
5 2
12)[Đ81III]
) 2 4 ( cos 2 sin 2 cos sin
2
sin
x x x
13)[ĐHQGHN _98] sin3x + cos3x = 2(sin5x + cos5x)
14)[ĐHNT_00]
sin8x + cos8x = 2(sin10x + cos10x) +
4
5
cos2x 15)[HVCNBCVT_98] sin4x –cos4x = 1 + 4(sinx –
cosx)
16)[Đ97II] 6tgx = tg2x
17)[HVNH TPHCM_98] 2 + cosx =
2
2tg x
18)[ĐHTC_97] (1 – tgx)(1 + sin2x) = 1 + tgx
19)[ĐHM_99] tgxsin2x – 2sin2x = 3(cos2x +
sinxcosx)
20)[ĐHQGHN_D00] 1 + 3tgx = 2sin2x
21)[viện ĐH Mở HN_98] cosx +
2
x
tg = 1 22)[ĐH Dân Lập Đông Đô_99] cotgx = tgx + 2tg2x
Chuyên Đề 11: sử dụng CT góc nhân ba
1)[ĐHY Hải Phòng_00] sin3x + sin2x = 5sinx
2) sin3x + sinx – 2cos2x = 0
3)[ĐHY Thái Nguyên] 4cos2x–
cos3x=6cosx+2(1+cos2x)
4)[Đ76II]
cos10x+2cos24x+6cos3xcosx = cosx +8cosxcos33x
5) 32cos6x – cos6x = 1
6)[ĐHTH_B92] 2sin3x(1 – 4sin2x) = 1
7)[ĐHQGHN_01]
sin3x = cosxcos2x(tg2x + tg2x)
8)[ĐHTM_99]
x
x x
x
cos
1 3 cos 2 sin
1 3 sin
2
3 10
sin(
2
1 ) 2 10
3 sin( π −x = π + x
4 sin(
2 sin ) 4 3 sin( x−π = x x+π
11)[ĐHQG_99] x ) cos 3x
3 ( cos
8 3 +π = 12)[HVNH TPHCM_00]
sin3x + cos3x + 2cosx = 0
Chuyên Đề 12: biến đổi tổng, hiệu thành tích và phân tích ra thừa số
1) sinx + sin2x + sin3x = 1 + cosx + cos2x 2) sinx + sin2x + sin3x = cosx + cos2x + cos3x 3)[ĐH Nông Lâm TPHCM_01]
1 + cosx + cos2x + cos3x = 0 4)[HVQHQT_99] cosx + cos2x + cos3x + cos4x = 0 5)[ĐHSP Vinh_97]
sinx + sin2x + sin3x + sin4x + sin5x + sin6x = 0 6)[ĐH Đà Nẵng_B97] sin3x – sinx + sin2x = 0
7) cos10x – cos8x – cos6x + 1 = 0 8)[HVQHQT_00] cosx + cos3x + 2cos5x = 0 9)[ĐHNTHN_97] 9sinx + 6cosx – 3sin2x + cos2x = 8 10)[ĐHNT TPHCM_00]
1 + sinx + cos3x = cosx + sin2x + cos2x 11)[ĐHYHN_00] sin4x = tgx
12) (2sinx – 1)(2sin2x + 1) = 3 – 4cos2x 13)[ĐHYHN_96] (cosx – sinx)cosxsinx = cosxcos2x 14)[ĐHHH_00]
(2sinx + 1)(3cos4x + 2sinx – 4) + 4cos2-x = 3 15)[ĐH Đà Nẵng_99] cos3x – sin3x = sinx – cosx 16)[ĐH Thuỷ Sản Nha Trang_96]
cos3x + sin3x = sinx – cosx 17)[ĐHCSND_00] cos3x + sin3x = sin2x + sinx +cosx 18)[HVQY_00] cos2x + sin3x + cosx = 0
19)[HVNH_99] cos3x + cos2x + 2sinx – 2 = 0 20)[HVNH TPHCM_00] sinx + sin2x + cos3x = 0 21)[HVBCVT TPHCM_97] cos2x – 4sinxcosx = 0 22)[HVKTQS_99] 2sin3x – sinx =2cos3x –cosx + cos2x
23)[ĐHSPI_00] 4cos3x +3 2sin2x = 8cosx 24)[ĐHNTHN_98]
sinx +sin2x +sin3x +sin4x =cosx+cos2x+cos3x+cos4x 25)[ĐH Thuỷ Sản Nha Trang_97]
x x
x
2 sin 2
sin 2 cos 4 − 4 = 26)[HVQY_97]
0 1 2 sin ) 3 (sin 2 sin ) 3 (sinx+ 4 x− x+ 2 x+ = 27)[ĐHQGHN_B97]
x x
x
cos
1 sin
1 ) 4 sin(
2
28)[ĐHKTHN_98]
x x
2 2
sin
1 cos
29)[ĐHTL_00] 5sin3x = 3sin5x
Trang 430)[ĐHM_97] 1
sin 5
5 sin
=
x x
31)[ĐHNNHN_00] 2cos2x – 8cosx + 7 = 1/cosx
Chuyên Đề 13: sd CT biến đổi tích thành
tổng
1) cos11x.cos3x = cos17x.cos9x
2) sin18x.cos13x = sin9x.cos4x
3) sin2x + sin2xsin4x + sin3xsin9x + sin4xsin16x = 1
4) (sinx + 3cosx)sin3x = 2
6 sin(
) 6 sin(
cos
6)[ĐHBK TPHCM_91]
sin3x -
3
2
sin2x = 2sinxcos2x 7) 8sinx =
x
x sin
1 cos
3 + 8)[ĐHGT_96] cos3xtg5x = sin7x
9)
) 3
2 cos(
cos 3 4 ) 3 sin(
).
3
sin(
.
sin
) 2
3
4 cos(x+ π = 10)[ĐHTHHN_92] 2sin3x(1 – 4sin2x) = 1
11)[ĐHDHN_00]
cos2x + cos4x + cos6x = cosxcos2xcos3x + 2
12)[ĐHYHN_97]
2
1 2
3 sin 2 sin sin 2
3 cos 2 cos
13)[HVQHQT_96]
2 sin cos 5 2
5
x
x = 14)[ĐHTL TPHCM_00]
tgx – 3cotgx = 4(sinx + 3 cosx)
15)[ĐHKT_00]
x x
x ) 1 8 sin 2 cos 2
4 3
sin(
Chuyên Đề 14: PTLG phối hợp(tg,sin),
(cotg,cos)
1) 2(tgx – sinx) + 3(cotgx – cosx) + 5 = 0
2)[ĐHGT_97] 3(cotgx – cosx) – 5(tgx – sinx) = 2
3)[ĐHDL Hồng Đức Thanh Hoá_99]
4sin2x + 3tg2x = 1
4)[ĐHM_99] 1 + tgx = 2 2 sinx
5)[ĐHQG_96] 1 + 3sin2x = 2tgx
6)[ĐHQGHN_95] tg2x(1 – sin3x) + cos3x – 1 = 0
7)[ĐHQG_A00] 2sinx + cotgx = 2sin2x + 1
8)ĐH Nông Lâm TPHCM_97]
x
x x
tg
sin 1
cos 1 2
−
+
= 9)[ĐH Thuỷ Sản Nha Trang_97]
x
x x
g
cos 1
sin 1 cot 2
+
+
= 10) tg2x = 11−−cossinx x
11)[CĐHQ_96] tg2x =
x
x
3
3
sin 1
cos 1
−
−
12) tg2x =
x
x
3 3
sin 1
cos 1
+ +
tgx
tgx
2 sin 1 1
1
+
= +
−
14)[ĐHSP Vinh_98] 1 + cotg2x =
x
x
2 sin
2 cos 1
2
−
15)[Đ56II] Tìm tổng các nghiệm x ∈ [ 1 ; 70 ] của PT cos2x – tg2x =
x
x x
2
3 2
cos
1 cos
16)[Đ140II] Tìm tổng các nghiệm x ∈ [ 2 ; 40 ] của PT 2cos2x + cotg2x =
x
x
2
3
sin
1 sin +
Chuyên Đề 15: PTLG dạng phân thức
1)[Đ30II]
x x
2 2
sin
1 cos
1 sin cos
2
cos sin 2 cos
+ +
−
x x
x x x
9 cos
5 cot
x
x g x
4)[Đ119II]
x
x x
x
4 cos 1
4 sin 2
sin 2
4 cos 1
+
=
−
3 cos 2 cos cos
3 sin 2 sin
+ +
+ +
x x
x
x x
x
2 sin ) cos (sin
2 cos 4 cos sin
2 2
=
−
− +
x x
x
x x
x
1 cos sin 2
2 sin sin 2 3 sin 2
=
−
+
− +
x x
x x
x
8)[Đ2II]
3
10 sin
1 sin
cos
1
x
x x x
1 sin 4 cos 3
6 sin
4 cos
+ +
+ +
x x
x x
10)[ĐHSP Vinh_98] 1 + cotg2x =
x
x
2 sin
2 cos 1
2
−
11) tg3x.cotgx = -1