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
) – cos23x – 3cos(
2
- 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
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
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 3 ( 6 2 3 ) cos 2 ( 4 3 3 ) cos 2 3 0
x
4) cos3x + 3cos2x = 2(1 + cosx)
5) 2tg3x + 5tg2x – 23tgx + 10 = 0
6) 6 3 ( 3 2 3 ) 2 ( 3 3 ) 3 0
x
tg
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
cos7x 3sin7x 2
5)[ĐHGT_00] 2 2 (sinx cosx) cosx 3 cos 2x
6 5 sin(
5 ) 6 sin(
4 ) 3 sin(
7)[HVCNBCVT_01]
3 4 cos 3 3 3 sin cos 4 3 cos
sin
x x
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
) 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
12)[ĐHQGHN_98] x ) cos 3x
3 ( cos
13)[ĐHQG TPHCM_98] x ) 2 sinx
4 (
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
9) 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
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
Trang 2
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
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 (
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
12)[ĐHD_99] sin24x – cos26x = sin( 10 , 5 10x)
13)[ĐHTDTT_01]
cos3x+sin7x=
2
9 cos 2 ) 2
5 4 ( sin
14)[ĐHHH_95] sin4x +
4
1 ) 4 (
x
15)[ĐHBKHN_95]
2sin2x(4sin4x – 1) = cos2x(7cos22x + 3cos2x – 4) 16)[ĐHXD_97]
x x
tg x tg
x
) 4 ( ) 4 (
2 cos 2
11)[ĐHGT_99]
6 ( cot ) 3 ( cot 8
7
x g x
12)[ĐHGT_01]
Sin4x +
8
9 ) 4 ( sin ) 4 (
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
19)[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
x x
2
7 cos(
3 ) 2
5 2
12)[Đ81III]
) 2 4 ( cos 2 sin 2 cos sin
2 sin
x
x x
x
Trang 3
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
4 sin(
2 sin ) 4 3
11)[ĐHQG_99] x ) cos 3x
3 ( cos
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
26)[HVQY_97]
0 1 2 sin ) 3 (sin 2 sin ) 3
x
x x
27)[ĐHQGHN_B97]
x x
x
cos
1 sin
1 ) 4 sin(
2
28)[ĐHKTHN_98]
x x
2 2
sin
1 cos
1
29)[ĐHTL_00] 5sin3x = 3sin5x
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
x
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
x x
x x
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
Trang 4
7)[§HQG_A00] 2sinx + cotgx = 2sin2x + 1
8)§H N«ng L©m TPHCM_97]
x
x x
tg
sin 1
cos 1
2
9)[§H Thuû S¶n Nha Trang_97]
x
x x
g
cos 1
sin 1
10) tg2x =
x
x
sin 1
cos 1
11)[C§HQ_96] tg2x =
x
x
3
3
sin 1
cos 1
12) tg2x =
x
x
3
3
sin 1
cos 1
tgx
tgx 1 sin2
1
1
14)[§HSP Vinh_98] 1 + cotg2x =
x
x
2 sin
2 cos 1
2
15)[§56II] T×m tæng c¸c nghiÖm x [ 1 ; 70 ] cña PT
cos2x – tg2x =
x
x x
2
3 2
cos
1 cos
16)[§140II] T×m tæng c¸c nghiÖm x [ 2 ; 40 ] cña PT
2cos2x + cotg2x =
x
x
2
3
sin
1 sin
Chuyªn §Ò 15: PTLG d¹ng ph©n thøc
1)[§30II]
x x
2 2
sin
1 cos
1
1 sin cos
2
cos sin 2 cos
x x
x x x
9 cos
5 cot sin
x
x g x
4)[§119II]
x
x x
x
4 cos 1
4 sin 2
sin 2
4 cos 1
3 cos 2 cos cos
3 sin 2 sin sin
x x
x
x x
x
2 sin ) cos (sin
2 cos 4
cos sin
x x
x
x x
x
1 cos sin 2
2 sin sin 2 3 sin 2
x x
x x
x
8)[§2II]
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)[§HSP Vinh_98] 1 + cotg2x =
x
x
2 sin
2 cos 1
2
11) tg3x.cotgx = -1