Gv: Trần Minh Hùng PHƯƠNG TRÌNH LƯỢNG GIÁC. (Tổng hợp) 1/ cos 2 3x.cos2x – cos 2 x = 0 2/ 1 + sinx + cosx + sin2x + cos2x = 0 3/ cos 4 x + sin 4 x + cos . 4 − π x sin − 4 3 π x - 2 3 = 0 4/ 5sinx – 2 = 3(1 – sinx)tan 2 x 5/ (2cosx – 1)(2sinx + cosx) = sin2x – sinx. 6/ cotx – 1 = 2 1 sin tan1 2cos 2 −+ + x x x sin2x. 7/ cotx – tanx + 4sin2x = x2sin 2 8/ 0 2 costan. 42 sin 222 =− − x x x π 9/ 32cos 2sin21 3sin3cos sin5 += + + + x x xx x với 0 < x < 2 π 10/ sin 2 3x – cos 2 4x = sin 2 5x – cos 2 6x 11/ cos3x – 4cos2x + 3cosx – 4 = 0 với 0 ≤≤ x 14 12/ cosx + cos2x + cos3x = sinx + sin2x + sin3x 13/ 26sin.222sin.3 2 −=− xx . 14/ cos3x + sin7x = 2. 2 9 cos2 2 5 4 sin 22 xx − + π 15/ sin 3 x + sinx.cosx = 1 – cos 3 x 16/ 2 + cos2x = 2tanx 17/ sinx.cosx + cos 2 x = 2 12 + 18/ −= + 24 sin.3 42 3 sin xx ππ 19/ sin3x + cos2x =2 ( sin2x.cosx – 1) 20/ 4cosx – 2cos 2 x – cos2x – cos4x = 0 21/ 1 2cos1 2sin = + + x x 22/ cosx + sin2x = 0 23/ 2(cos 4 x – sin 4 x) + cos4x – cos2x = 0 24/ (5sinx – 2)cos 2 x = 3(1 – sinx)sin 2 x 25/ (2sinx – 1)(2cosx + sinx) = sin2x – cosx 26/ cos3x + 2cos2x = 1 – 2sinxsin2x 27/ += ++ + 4 cos 6 cos 3 cos πππ xxx 28/ sin 3 x + cos 3 x = sinx – cosx 29/ xxx tansin.2 4 sin.2 22 −= − π 30/ 4cos 2 x – 2cos 2 2x = 1 + cos4x 31/ cos3x.sìnx – cos4x.sinx = xx cos13sin 2 1 ++ . 32/ (2sinx – 1)(2cos2x + 2sinx + 3) = 4sin 2 x – 1 33/ cosx.cos7x = cos3x.cos5x 34/ 3 2coscos 2sinsin = − − xx xx 35/ sinx + sin2x + sin3x = 0 36/ x xx xx 2tan 8 13 sincos sincos 22 66 = − + 37/ cos 2 x.sin 4 x + cos 2x = 2cosx(sinx + cosx) – 1 38/ 3 – tanx(tanx + 2sinx) + 6cosx = 0 39/ cos2x + cosx(2tan 2 x – 1) = 2 40/ 3cos4x – 8cos 6 x + 2cos 2 x + 3 = 0 41/ 1cos2 42 sin2cos)32( 2 − −−− x x x π = 1 42/ )sin1(2 cossin )1(coscos 2 x xx xx += + − 43/ cotx = tanx + x x 2sin 4cos2 44/ x x x xx 2sin.8 1 2cot 2 1 2sin.5 cossin 44 −= + 45/ x xx x 4 2 4 cos 3sin)2sin2( 1tan − =+ Gv: Trần Minh Hùng 46/ tanx + cosx – cos 2 x = sinx(1 + tanx.tan ) 2 x 47/ sin( 1)cos. =x π 48/ cos3x – sìnx = 3 (cos2x - sin3x) 49/ 2cos 2 x - sin2x + sinx – cosx = 0 50/ sin3x + cos2x = 1 + sinx.cos2x 51/ 1 + sinx + cosx + sin2x + cos2x = 0 52/ cos2x + 5sinx + 2 = 0 53/ cos 2 x.sin 2 x + cos2x = 2(sinx + cosx)cosx – 1 54/ 8.sin 2 x + cosx = 3 .sinx + cosx 55/ 3cos2x + 4cos 3 x – cos3x = 0 56/ 1 + cosx – cos2x = sinx + sin2x 57/ sin4x.sin2x + sin9x.sin3x = cos 2 x 58/ 0cossin1 =++ xx 59/ ( ) 1sin.sin22cossin1cos3 2 −=−− xxxxx 60/ 2cos.3 2 cos 2 sin 2 =+ + x xx 61/ −= − + x x x 4 7 sin4 2 3 sin 1 sin 1 π π 62/ 2sin 2 2x + sin7x – 1 = sinx 63/ 0 sin22 cossin)sin(cos2 66 = − −+ x xxxx 64/ cotx + sinx 4 2 tan.tan1 = + x x 65/ cos3x + cos2x – cosx – 1 = 0 Gv: Trần Minh Hùng HƯỚNG DẪN GIẢI 1/(1) ⇔ (1 + cos6x)(cos2x – (1 + cos2x) = 0 ⇔ cos6x.cos2x – 1 = 0 ⇔ cos8x + cos4x – 2 = 0 ⇔ 2cos 2 4x + cos4x – 2 = 0 ĐS: x = 2 π k 2/ (2) ⇔ sinx + cosx + 2sinxcosx + 2cos 2 x = 0 ⇔ sinx + cosx + 2cosx(sinx + cosx) = 0 ⇔ (sinx + cosx)(2cosx + 1) = 0 Đ S: x = π π π π 2 3 2 4 kxk +±=∨+− 3/ (3) 0 2 3 2sin 2 4sin 2 1 cossin21 22 =− + −+−⇔ xxxx π 032sin4cos2sin2 2 =−+−−⇔ xxx 012sin)2sin21(2sin 22 =−+−−−⇔ xxx 022sin2sin 2 =−+⇔ xx ĐS: π π kx += 4 4/ Đk: π π kxx +≠⇔≠ 2 0cos (4) 02sin3sin2)sin1( sin1 sin3 2sin5 2 2 2 =−+⇔− − =−⇔ xxx x x x ĐS: π π π π 2 6 5 2 6 kxkx +=∨+= 5/ (5) 0)cos)(sin1cos2( =+−⇔ xxx ĐS: π π π π kxkx + − =∨+±= 4 2 3 6/ Đk: −≠ ≠ ≠ 1tan 0cos 0sin x x x (*) (6) )cos(sinsin)sin(coscos sin sincos )cos(sinsin cos sin 1 sincos 1 sin cos 22 xxxxxx x xx xxx x x xx x x −+−= − ⇔−+ + − =−⇔ =+− =− ⇔=+−−⇔ 0sincossin1 0sincos 0)sincossin1)(sin(cos 2 2 xxx xx xxxxx . ĐS: π π kx += 4 7/ Đk: ≠ ≠ 0cos 0sin x x (*) (7) x x x x x x 2sin 2 2sin4 cos sin sin cos =+−⇔ 22sin42cos2 2sin 2 2sin4 cossin sincos 2 22 =+⇔=+ − ⇔ xx x x xx xx 012cos2cos2 2 =−−⇔ xx +±= = ⇔ −= = ⇔ π π π kx kx x x 3 2 1 2cos 12cos Kết hợp với Đk (*), nghiệm của (7) là x = π π k+± 3 8/ Đk: 0cos ≠x (*) (8) xxxxx x x x 22 2 2 cos)cos1(sin)sin1()cos1( 2 1 cos sin 2 cos1 2 1 +=−⇔+= −−⇔ π Gv: Trần Minh Hùng +−= += += ⇔ −= −= = ⇔=++−⇔ π π ππ π π kx kx kx x x x xxxx 4 2 2 2 1tan 1cos 1sin 0)cos)(sincos1)(sin1( Kết hợp với Đk (*) , nghiệm của (8) là: π π ππ kxkx +−=∨+= 4 2 9/ Đk: 2 1 2sin −≠x (*) (9) 32cos 2sin21 3sin3cos3coscossin 532cos 2sin21 3sin3cos2sinsin2sin 5 += + ++−+ ⇔+= + +++ ⇔ x x xxxxx x x xxxxx 2 1 cos02cos5cos232coscos532cos 2sin21 cos)12sin2( 5 2 =⇔=+−⇔+=⇔+= + + ⇔ xxxxxx x xx π π 2 3 kx +±=⇔ Vì )2;0( π ∈x , ta có 3 5 , 3 ππ == xx thỏa (*) 10/ (10) 0)7cos11(coscos 2 12cos1 2 10cos1 2 8cos1 2 6cos1 =−⇔ + − − = + − − ⇔ xxx xxxx = = ⇔=⇔=⇔ 2 9 02sin.9sin02sin9sincos π π k x k x xxxxx 11/ (11) ⇔ ( cos3x + 3cosx) – 4(cos2x + 1) = 0 ⇔ 4cos 3 x – 8cos 2 x = 0 ⇔ 4cos 2 x(cosx- 2) = 0 ⇔ cosx = 0 π π kx +=⇔ 2 [ ] 3,2,1,014;0 ====⇔∈ kkkkx . ĐS: 2 7 , 2 5 , 2 3 , 2 ππππ ==== xxxx 12/ (12) ⇔ (2cosx + 1)(sin2x – cos2x) = 0 ⇔ 2cosx + 1 = 0 hoặc sin2x – cos2x = 0 ĐS: 28 ,2 3 2 ππ π π k xkx +=+±= 13/ (13) 62cos22sin3 =+⇔ xx . ĐS : VN 14/ (14) ⇔ cos3x + sin 7x = 1 - )19(cos5 2 cos +− + xx π ⇔ cos3x + cos9x + sin7x - sin5x = 0 ⇔ 2cos6x(cos3x + sinx) = 0. ĐS: 28 , 4 , 612 ππ π πππ k xkx k x +−=+=+= 15/ (15) ⇔ sin 3 x + sinx.cosx + cos 3 x = 1 ⇔ (sinx + cosx)(sin 2 x –sinx.cosx + cos 2 x) + (sinxcosx -1) = 0 ⇔ (sinxcosx – 1)(1 – sinx – cosx) = 0 ĐS: ππ π 2,2 2 kxkx =+= 16/ ĐK: cosx 0≠ Gv: Trần Minh Hùng (16) 0)3tantan2)(1(tan03tan2tantan2tan2 tan1 tan1 2 223 2 2 =++−⇔=−+−⇔= + − +⇔ xxxxxxx x x ĐS: π π kx += 4 17/ (17) 22cos2sin 2 12 )2cos1( 2 1 2sin 2 1 =+⇔ + =++⇔ xxxx . ĐS: π π kx += 8 18/ Đặt : t x t x t x 3 42 3 3 4 3 2 3 24 −=+⇔−=⇔=− π πππ (18) 2 4 0sinsin3sin4sin3sin33sin 3 ππ π k xkttttttt − =⇔=⇔=⇔=−⇔=⇔ 19/ (19) ⇔ sin 3x + 1 – 2sin 2 x = sin3x + sinx - 2 ⇔ (2sinx + 3)(sinx – 1) = 0. ĐS: π π 2 2 kx += 20/ (20) ⇔ 4cosx – 2cos 2 x – 2cos3xcosx = 0 ⇔ 2cosx(2 – cosx – cos3x) = 0 ĐS: ππ π 2, 2 kxkx =+= 21/ ĐK: cosx 1−≠ ĐS: π π π π π 2 6 7 ,2 6 , lxlxkx +=+−== 22/ ĐS: 3 2 2 ,2 2 ππ π π k xkx +=+−= 23/ (23) ⇔ 2cos2x + 2cos 2 2x – 1 – cos2x = 0 ⇔ 2cos 2 2x + cos2x – 1 = 0. ĐS: π π π π kxkx +±=+= 6 , 2 24/ (24) ⇔ 5sinx(1 – sin 2 x) – 2(cos 2 x + sin 2 x) – sin 2 x + 3sin 3 x = 0 ⇔ 2sin 3 x + sin 2 x – 5sinx + 2 = 0. ĐS: π π π π π π 2 6 5 ,2 6 ,2 2 kxkxkx +=+=+= 25/ (25) ⇔ (2sinx – 1)(2cosx + sinx) = 2sinxcosx – cosx ⇔ (2sinx – 1)(cosx + sinx) = 0. ĐS: π π π π π π kxkxkx +−=+=+= 4 ,2 6 5 ,2 6 26/ (26) ⇔ cos3x + 2cos2x = 1- cosx + cos3x ⇔ 2(2cos 2 x – 1) = 1 – cosx ⇔ 4cos 2 x + cosx – 3 = 0. ĐS: παππ 2,2 kxkx +±=+= 27/ (27) 0 4 cos01 12 cos2 4 cos 4 cos 12 cos 4 cos2 = +⇔= − +⇔ += +⇔ ππππππ xxxx 28/ (28) )sincos.sincos2(cos)cos)(sincos(sincossin 222233 xxxxxxxxxxx +−⇔+−=+⇔ + cosx = 0 + ))(0(cos02tantan0sincos.sincos2 22 VNptnghiemlakhongxxxxxxx ==+−⇔=+− 29/ ĐK : π π kxx +≠⇔≠ 2 0cos (29) )cos(sinsin2 cos cossin tansin221tansin2 2 2cos1 22 xxx x xx xxxsìnxxx += + ⇔−=−⇔−= −−⇔ π 0)2sin1)(cos(sin =−+⇔ xxx 30/ (30) xxxxxxxxx cos2coscos2cos2coscos2cos22cos2cos4 22222 −=∨=⇔=⇔=−⇔ 31/ (31) xxxxxxxx cos12sincos13sin 2 1 )3sin5(sin 2 1 )sin5(sin 2 1 +=−⇔++=−−−⇔ Gv: Trần Minh Hùng ππ 2 1cos 0sin 3cos1cos 0sin )cos1(4sin 0sin 2 kx x x xx x xx x +=⇔ −= ≤ ⇔ −=∨−= ≤ ⇔ += ≤ ⇔ 32/ (32) 0)sin44)(1sin2( 2 =−−⇔ xx 33/ (33) 442 2cos6cos πππ k x k x k xxx =⇔=∨=⇔=⇔ 34/ ĐK: 3 2 202coscos π π k xkxxx ≠∧≠⇔≠− (34) xxxxxxxx sin 2 1 cos 2 3 2sin 2 1 2cos 2 3 2cos3cos32sinsin −=−⇔−=−⇔ 3 2 9 2 6 cos 6 2cos ππ π ππ k xkxxx +−=∨=⇔ += +⇔ 35/ (35) 2 1 cos02sin0cos.2sin22sin −=∨=⇔=+⇔ xxxxx 36/ Đk: cos2x 24 0 ππ k x +≠⇔≠ (36) xxxxxxxxxx 2sin13)]cos(sincossin3)sin[(cos82sin13)sin(cos8 222232266 =+−+⇔=+⇔ )( 3 8 2sin 2 1 2sin082sin132sin62sin132sin 4 3 18 22 loaixxxxxx =∨=⇔=−+⇔= −⇔ 37/ Pt )0sincos2(02sin0)sincos2(cos.sin2sincos2cos2sin.cos 332242 ≠−=⇔=−⇔+=+⇔ xxxxxxxxxxxx 38/ ĐK: cosx 0≠ Pt 0cos6)cos21(sincos30cos6 cos cossin2sin cos sin 3 322 =++−⇔=+ + −⇔ xxxxx x xxx x x 0)sincos3)(cos21(0)cos21(sin)cos21(cos3 2222 =−+⇔=+−+⇔ xxxxxxx π π kxxxx x x +±=⇔−=⇔=+⇔=⇔ = −= ⇔ 32 1 2cos 2 1 2cos1 4 1 cos 4 1 cos 2 1 cos 2 2 39/ ĐK : cosx 0≠ Pt x x xxxx x x x x x x cos11 cos 1 sin2sin212cos2cos cos sin 22cos cos sin 22cos 22 22 += −⇔+=−=−⇔=−+⇔ 0]cos)cos1(2)[cos1(cos)cos1()cos1)(cos1(2 22 =−−+⇔+=−−⇔ xxxxxxx = −= ⇔ =+− −= ⇔ 2 1 cos 1cos 02cos5cos2 1cos 2 x x xx x 40/ Pt 0)1cos2)(1cos2(cos22cos60)1cos4(cos2)4cos1(3 222242 =−+−⇔=−−+⇔ xxxxxxx 0)]1cos2(cos2cos3[2cos02cos)1cos2(cos22cos6 22222 =+−⇔=+−⇔ xxxxxxxx Gv: Trần Minh Hùng = =⇔=⇔= ⇔=−+−⇔=−−− +=⇔= ⇔ )( 2 3 cos 0sin1cos 03cos5cos20coscos2)1cos2(3 24 02cos 2 2 24242 loaix kxxx xxxxx k xx π ππ 41/ ĐK: 2 1 cos ≠x Pt 0cos 2 3 sin 2 1 20sincos31cos2 2 cos1cos)32( = −⇔=+−⇔−= −−−−⇔ xxxxxxx π π π π π π ππ )12( 3 2 1 cos 3 3 0 3 sin2 ++=⇔ ≠ += ⇔=−⇔= −⇔ nx x kx kxx 42/ ĐK: 0 4 sin2cossin ≠ +=+ π xxx Pt 0)]cos(sin2)1)(cossin1)[(sin1()sin1)(cos(sin2)1)(cossin1( 2 =+−−−+⇔++=−−⇔ xxxxxxxxxx )( 2 2 2 1cos 1sin 0)sin1)(cos1( 1sin đkthoa kx kx x x xx x += +−= ⇔ −= −= ⇔ =++ −= ⇔ ππ π π 43/ ĐK: 12cos02sin ≠⇔≠ xx Pt +±=⇔−= = ⇔ =−−⇔=⇔=−⇔ π π kxx loaix xxxx xx x x x x x 32 1 2cos )(12cos 012cos2cos24cos2cos cos.sin 4cos cos sin sin cos 2 44/ ĐK: sìn 2x 0 ≠ Pt 0 4 9 2cos52cos 8 1 2cos 2 1 5 2sin 2 1 1 8 1 2cos 2 1 5 cos.sin21 2 2 22 =+−⇔−= − ⇔−= − ⇔ xxx x x xx +±=⇔= = ⇔ π π kxx loaix 62 1 2cos )( 2 9 2cos 45/ ĐK: 1sin0cos ≠⇔≠ xx Pt xxxxxxxxxx 3sin)2sin2(2sin23sin)2sin2(2sin 2 1 13sin)2sin2(cossin 2222244 −=−⇔−=−⇔−=+⇔ 2 1 3sin =⇔ x Gv: Trần Minh Hùng 46/ ĐK: ≠ ≠ 0 2 cos 0cos x x Ta có: x x x x x x x x x x x x x x x x cos 1 2 cos.cos 2 cos 2 cos.cos 2 sin.sin 2 cos.cos 2 cos.cos 2 sin.sin 1 2 tan.tan1 == + =+=+ Pt π 2)0(cos1cos0)cos1(cos cos sin coscostan 2 kxxxxx x x xxx =⇔≠=⇔=−⇔=−+⇔ . = −= ⇔ =+− −= ⇔ 2 1 cos 1cos 02cos5cos2 1cos 2 x x xx x 40/ Pt 0)1cos2)(1cos2(cos22cos60)1cos4(cos2)4cos1(3 222242 =−+−⇔=−−+⇔ xxxxxxx 0)]1cos2(cos2cos3[2cos02cos)1cos2(cos22cos6 22222 =+−⇔=+−⇔. ≠ ≠ 0 2 cos 0cos x x Ta có: x x x x x x x x x x x x x x x x cos 1 2 cos.cos 2 cos 2 cos.cos 2 sin.sin 2 cos.cos 2 cos.cos 2 sin.sin 1 2 tan.tan1 == + =+=+ Pt π 2)0(cos1cos0)cos1(cos cos sin coscostan 2 kxxxxx x x xxx. +±=⇔−= = ⇔ =−−⇔=⇔=−⇔ π π kxx loaix xxxx xx x x x x x 32 1 2cos )(12cos 012cos2cos24cos2cos cos.sin 4cos cos sin sin cos 2 44/ ĐK: sìn 2x 0 ≠ Pt 0 4 9 2cos52cos 8 1 2cos 2 1 5 2sin 2 1 1 8 1 2cos 2 1 5 cos.sin21 2 2 22 =+−⇔−= − ⇔−= − ⇔