Đăng ký Đăng nhập
Trang chủ Giáo dục - Đào tạo Trung học phổ thông Lớp 11 200 CÂU GIẢI PHƯƠNG TRÌNH LƯỢNG GIÁC...

Tài liệu 200 CÂU GIẢI PHƯƠNG TRÌNH LƯỢNG GIÁC

.DOC
5
570
91

Mô tả:

200 CÂU GIẢI PHƯƠNG TRÌNH LƯỢNG GIÁC
........ Các dạng bài tập lượng giác a/kiÕn thøc cÇn nhí vµ ph©n lo¹i bµi to¸n d¹ng 1 Ph¬ng tr×nh bËc nhÊt vµ bËc hai , bËc cao víi 1 hµm sè lîng gi¸c §Æt HSLG theo t víi sinx , cosx cã ®iÒu kiÖn t �1 Gi¶i ph¬ng tr×nh ……….theo t NhËn t tho¶ m·n ®iÒu kiÖn gi¶i Pt lîng gi¸c c¬ b¶n Gi¶i ph¬ng tr×nh: 2cos2x- 4cosx=1 � 1/ � 2/ 4sin3x+3 2 sin2x=8sinx sinx � 0 � 1-5sinx+2cosx=0 � 3/ 4cosx.cos2x +1=0 4/ � � cos x �0 5/ Cho 3sin3x-3cos2x+4sinx-cos2x+2=0 (1) vµ cos2x+3cosx(sin2x-8sinx)=0 (2). 1 T×m n0 cña (1) ®ång thêi lµ n0 cña (2) ( nghiÖm chung sinx= ) 3 3 4 6/ sin3x+2cos2x-2=0 7/ a/ tanx+ -2 = 0 b/ +tanx=7 cot x cos 2 x c* / sin6x+cos4x=cos2x 5 7 8/sin( 2 x  )-3cos( x  )=1+2sinx 9/ sin 2 x  2sin x  2  2sin x  1 2 2 10/ cos2x+5sinx+2=0 11/ tanx+cotx=4 13/ sin x  1  cos x  0 2 4 15/ 4sin 2 x  6sin x  9  3cos 2 x  0 cos x d¹ng 2: 12/ 14/ cos2x+3cosx+2=0 b a 2  b2 � a  b sin( x   )  c 2 C¸ch 3: §Æt §¨c biÖt : 16/ 2cosx- sin x =1 Ph¬ng tr×nh bËc nhÊt ®èi víi sinx vµ cosx : asinx+bcosx=c C¸ch 1: asinx+bcosx=c a §Æt cosx= 2 ; sinx= a  b2 2 sin 2 2 x  4 cos 4 2 x  1 0 2sin x cos x t  tan C¸ch : 2 b � � a� sin x  cos x � c a � � b  tan  � a  sin x  cos x.tan    c a c � sin( x   )  cos  a §Æt 2 x ta cã sin x  2t 2 ;cos x  1  t 2 � (b  c)t 2  2at  b  c  0 2 1 t 1 t   sin x  3 cos x  2sin( x  )  2 cos( x  ) 3 6   2. sin x �cos x  2 sin( x � )  2 cos( x m ) 4 4   3. sin x  3 cos x  2sin( x  )  2 cos( x  ) 3 6 2 2 2 §iÒu kiÖn Pt cã nghiÖm : a  b �c 1. gi¶i ph¬ng tr×nh : 1/ 2sin15x+ 3 cos5x+sin5x=k 2/ a : 3 sin x  cos x  c: 3/ 1 cos x 3 sin x  cos x  3  víi k=0 b: 4sin x  3cos x  1 3 sin x  cos x  1 cos 7 x  3 sin 7 x  2  0 Chuyªn ®Ò ph¬ng trinh lîng gi¸c víi k=0 vµ k=4 6 6 4sin x  3cos x  1 *t×m nghiÖm x �( 2 6 ; ) 5 7 1 ........ 4/( cos2x- 3 sin2x)6/ 3 sinx-cosx+4=0 5/ cos x  2sin x.cos x  3 2 cos 2 x  sin x  1 1  cos x  cos 2 x  cos 3x 2  (3  3 sin x) 2 cos 2 x  cos x  1 3 D¹ng 3 Ph¬ng tr×nh ®¼ng cÊp ®èi víi sin x vµ cosx §¼ng cÊp bËc 2: asin2x+bsinx.cosx+c cos2x=0 C¸ch 1: Thö víi cosx=0 Víi cosx �0 .Chia 2 vÕ cho cos2x ta ®îc: atan2x+btanx +c=d(tan2x+1) C¸ch2: ¸p dông c«ng thøc h¹ bËc §¼ng cÊp bËc 3: asin3x+b.cos3x+c(sinx+ cosx)=0 hoÆc asin3x+b.cos3x+csin2xcosx+dsinxcos2x=0 XÐt cos3x=0 vµ cosx �0 Chia 2 vÕ cho cos2x ta ®îc Pt bËc 3 ®èi víi tanx Gi¶i ph¬ng tr×nh 1/a/ 3sin2x- 3 sinxcosx+2cos2x cosx=2 b/ 4 sin2x+3 3 sinxcosx-2cos2x=4 c/3 sin2x+5 cos2x-2cos2x-4sin2x=0 d/ 2 sin2x+6sinxcosx+2(1+ 3 )cos2x-5- 3 =0 3 2/ sinx- 4sin x+cosx=0 2 c¸ch +/ (tanx -1)(3tan2x+2tanx+1)=0  x   k + sin3x- sinx+ cosx- sinx=0 � (cosx- sinx)(2sinxcosx+2sin2x+1)=0 4 2 2 3/ tanx sin x-2sin x=3(cos2x+sinxcosx) 4/ 3cos4x-4sin2xcos2x+sin4x=0 5/ 4cos3x+2sin3x-3sinx=0 6/ 2 cos3x= sin3x 7/ cos3x- sin3x= cosx+ sinx 3 8/ sinx sin2x+ sin3x=6 cos x 9/sin3(x-  /4)= 2 sinx Dang 4 Ph¬ng tr×nh vÕ tr¸i ®èi xøng ®èi víi sinx vµ cosx * a(sin x+cosx)+bsinxcosx=c ®Æt t= sin x+cosx t � 2 * t 2  1 =c � at + b � bt2+2at-2c-b=0 2 a(sin x- cosx)+bsinxcosx=c ®Æt t= sin x- cosx t � 2 1  t =c � at + b � bt2 -2at+2c-b=0 2 2 Gi¶i ph¬ng tr×nh 1 1 1 1/ a/1+tanx=2sinx + b/ sin x+cosx= cos x tan x cot x 2/ sin3x+cos3x=2sinxcosx+sin x+cosx 3/ 1- sin3x+cos3x= sin2x 4/ 2sinx+cotx=2 sin2x+1 5/ 2 sin2x(sin x+cosx)=2 6/ (1+sin x)(1+cosx)=2 7/ 2 (sin x+cosx)=tanx+cotx 3 8/1+sin3 2x+cos32 x= sin 4x 9/* a* 3(cotx-cosx)-5(tanx-sin x)=2 2 9/b*: cos4x+sin4x-2(1-sin2xcos2x) sinxcosx-(sinx+cosx)=0 1 1 10 10/ sin x  cos x  4sin 2 x  1 11/ cosx+ +sinx+ = sin x 3 cos x 12/ sinxcosx+ sin x  cos x =1 dang 5 Gi¶i ph¬ng tr×nh b»ng ph¬ng ph¸p h¹ bËc C«ng thøc h¹ bËc 2 C«ng thøc h¹ bËc 3 1  cos 2 x 3cos x  cos 3 x 3sin x  sin 3 x 1  cos 2 x cos2x= ; sin2x= cos3x= ; sin3x= 2 4 2 4 Gi¶i ph¬ng tr×nh 1/ sin2 x+sin23x=cos22x+cos24x 2/ cos2x+cos22x+cos23x+cos24x=3/2 2 2 2 3/sin x+ sin 3x-3 cos 2x=0  5x 9x 4/ cos3x+ sin7x=2sin2(  )-2cos2 4 2 2 2 2 2 2 5/ sin 4 x+ sin 3x= cos 2x+ cos x víi x �(0;  ) Chuyªn ®Ò ph¬ng trinh lîng gi¸c 2 ........  6/sin24x-cos26x=sin( 10,5  10x ) víi x �(0; ) 7/ cos4x-5sin4x=1 2 8/4sin3x-1=3- 3 cos3x 9/ sin22x+ sin24x= sin26x 10/ sin2x= cos22x+ cos23x 11/ (sin22x+cos42x-1): sin x cos x =0  k   k � 2 2 2 ;  12/ 4sin3xcos3x+4cos3x sin3x+3 3 cos4x=3 x  � �  � 13/ 2cos 2x+ cos2x=4 sin 2xcos x 2 �24 2 8  x 14/ cos4xsinx- sin22x=4sin2(  )-7/2 víi x  1 <3 4 2 15/ 2 cos32x-4cos3xcos3x+cos6x-4sin3xsin3x=0  16/ sin3xcos3x +cos3xsin3x=sin34x 17/ * 8cos3(x+ )=cos3x 3 18/cos10x+2cos24x+6cos3xcosx=cosx+8cosxcos23x sin 5 x 19/ =1 5sin x 20 / cos7x+ sin22x= cos22x- cosx 21/ sin2x+ sin22x+ sin23x=3/2 22/ 3cos4x-2 cos23x=1 Dang 6 : Ph¬ng tr×nh LG gi¶i b»ng c¸c h»ng ®¼ng thøc * a3 �b3=(a �b)(a2 mab+b2) * a4- b4=( a2+ b2) ( a2- b2) * a8+ b8=( a4+ b4)2-2 a4b4 * a6 �b6=( a2 �b2)( a4 ma 2b2+b4) Gi¶i ph¬ng tr×nh 1/ sin4 x x +cos4 =1-2sinx 2 2 2/ cos3x-sin3x=cos2x-sin2x 3/ cos3x+ sin3x= cos2x 4/ sin 4 x  cos4 x 1  (tan x  cot x) v« nghiÖm sin 2 x 2 7   6/sin4x+cos4x= cot( x  ) cot(  x) 8 3 6 13 cos22x 8 7/ cos6x+sin6x=2(cos8x+sin8x) 8/cos3x+sin3x=cosx-sinx 9/ cos6x+sin6x=cos4x 10/ sinx+sin2x+sin3x+sin4x= cosx+cos2x+cos3x+cos4x x x 1 11/ cos8x+sin8x= 12/ (sinx+3)sin4 -(sinx+3) sin2 +1=0 8 2 2 Dang 7 : Ph¬ng tr×nh LG biÕn ®æi vÒ tÝch b»ng 0 1/ cos2x- cos8x+ cos4x=1 2/sinx+2cosx+cos2x-2sinxcosx=0 3/sin2x-cos2x=3sinx+cosx-2 4/sin3 x+2cosx-2+sin2 x=0 5/ 3sinx+2cosx=2+3tanx 6/ 3 sin2x+ 2 cos2x+ 6 cosx=0 2 sin 3 x sin 5 x 7/ 2sin2x-cos2x=7sinx+2cosx-4 8/  3 5 5 1 9/ 2cos2x-8cosx+7= 10/ cos8x+sin8x=2(cos10x+sin10x)+ cos2x cos x 4 11/ 1+ sinx+ cos3x= cosx+ sin2x+ cos2x 12/ 1+sinx+cosx+sin2x+cos2x=0 13/ sin2 x(tanx+1)=3sinx(cosx-sinx)+3 1 1 14/ 2sin3x=2cos3x+ 15/cos3x+cos2x+2sinx-2=0 sin x cos x 5/cos6x-sin6x= 16/cos2x-2cos3x+sinx=0 17/ tanx–sin2x-cos2x+2(2cosx- 18/sin2x=1+ 2 cosx+cos2x 20/ 2tanx+cot2x=2sin2x+ 22/ 1+tanx=sinx+cosx 24/ 2 2 sin( x  19/1+cot2x= 1 sin 2x  1 1 )=  4 sin x cos x Chuyªn ®Ò ph¬ng trinh lîng gi¸c 1  cos 2 x sin 2 2 x 1 )=0 cos x 21/cosx(cos4x+2)+ cos2x-cos3x=0 23/ (1-tanx)(1+sin2x)=1+tanx 25/ 2tanx+cotx= 3  2 sin 2x 3 ........ 26/ cotx-tanx=cosx+sinx 27/ 9sinx+6cosx-3sin2x+cos2x=8 Dang 8 : Ph¬ng tr×nh LG ph¶i thùc hiÖn c«ng thóc nh©n ®«i, h¹ bËc cos2x= cos2x- sin2x =2cos2x-1=1-2sin2x 2t 1 t2 sin2x=2sinxcosx sinx = ; cosx= 1 t2 1 t2 2 tan x tan2x= 1  tan 2 x Gi¶i ph¬ng tr×nh 1 1/ sin3xcosx= + cos3xsinx 2/ cosxcos2xcos4xcos8x=1/16 4 3/tanx+2cot2x=sin2x 4/sin2x(cotx+tan2x)=4cos2x 5/ sin4x=tanx 6/ sin2x+2tanx=3 7/ sin2x+cos2x+tanx=2 8/tanx+2cot2x=sin2x 9/ cotx=tanx+2cot2x 3 10/a* tan2x+sin2x= cotx b* (1+sinx)2= cosx 2 Dang 9 : Ph¬ng tr×nh LG ph¶i thùc hiÖn phÐp biÕn ®æi tæng_tÝch vµ tÝch_tæng Gi¶i ph¬ng tr×nh 1/ sin8x+ cos4x=1+2sin2xcos6x 2/cosx+cos2x+cos3x+cos4x=0 sin 3 x  sin x  sin 2 x  cos 2 x t×m x � 0; 2  4/ sinx+sin2x+sin3x+sin4x=0 3/ 1  cos 2 x 5/ sin5x+ sinx+2sin2x=1 6/ 3  cos 2 x  cot 2 x  cot 2 x  cos 2 x tanx= � � � �  4sin �  x � cos �  x � 4 4 � � � � 7/ tanx+ tan2x= tan3x 8/ 3cosx+cos2x- cos3x+1=2sinxsin2x Dang 10 : Ph¬ng tr×nh LG ph¶i ®Æt Èn phô gãc A hoÆc ®Æt hµm B Gi¶i ph¬ng tr×nh 3 x 1  3x x  �3  k 2 ; 4  k 2 ; 14  k 2 �   1/ sin( ) � )=sin2x sin( x  ) � 2/ sin( 3 x   )= sin(  15 15 �5 10 2 2 10 2 4 4 3  x 3/(cos4x/3 – cos2x): 1  tan 2 x =0 x  k 3 4/ cosx-2sin(  )=3 x  k 4 2 2 7  k � �  k ; 5/ cos( 2 x  )=sin(4x+3  ) x  � 6/3cot2x+2 2 sin2x=(2+3 2 )cosx � � 2 �6 2 7/2cot2x+ 2 +5tanx+5cotx+4=0 cos 2 x x   k 4 8/ cos2x+ 1 1 =cosx+ cos 2 x cos x 1  sin 2 x 1  tan x 1 1   7 �  k 2 �11/ +2 2 =5 x  � +2 =3 �  k 2 ;   k 2 ; 6 6 �2 sin x sin x 1  sin 2 x 1  tan x Dang 11 : Ph¬ng tr×nh LG ph¶i thùc hiÖn c¸c phÐp biÕn ®æi phøc t¹p 9/sinx- cos2x+ Gi¶i ph¬ng tr×nh 1/ 3  4 6  (16 3  8 2) cos x  4 cos x  3 3/ 5cos x  cos 2 x +2sinx=0 x   6  k 2 2  sin x  tan x  2 5/  2 cos x  2 x  � 3  k 2 tan x  sin x 7/tan2xtan23 xtan24x= tan2x-tan23 x+tan4x 9/sin3x=cosxcos2x(tan2x+tan2x)  � �4  � �  x  �  k 2 4  � x   k 4 2  � � x� �  k 2 ; �  k 2 � 4 �3 x  k x   k ;  k  , tan   2  � 2/cos � 3 x  9 x 2  16 x  80 �=1 t×m n0 x�Z x   21; 3 �4 � 4/3cotx- tanx(3-8cos2x)=0  x  �  k 3 6/sin3x+cos3x+ sin3xcotx+cos3xtanx= 2sin 2x x k 4  � tan 2 x � � 4 �  � x 2  � � x   k 2 4 k � 8/tanx+tan2x=-sin3xcos2x x  � �   k 2 � �3 10/ x  k 11/cos2 � sin x  2 cos 2 x �-1=tan2 � �x  x sin x  sin x  1  sin 2 x  cos x x  k ;sin x  5 1 2   k 2 4 x  � �x �3x  � � 6 sin �  � 2sin �  � 2sin �  � �5 12 � �5 12 � �5 3 � �5 6 � 12/ 2 cos � � 2t 1 t2 5 5 � 5 � x�   k 5 ;   k 5 ;  k 5 � 3 4 � 12 Dang 12 : Ph¬ng tr×nh LG kh«ng mÉu mùc, ®¸nh gi¸ 2 vÕ ,tæng 2 lîng kh«ng ©m,vÏ 2 ®å thÞ b»ng ®¹o hµm Gi¶i ph¬ng tr×nh 1/ cos3x+ 2  cos 2 3x =2(1+sin22x) x  k 2/ 2cosx+ 2 sin10x=3 2 +2sinxcos28x x  4  k 3/ cos24x+cos26x=sin212x+sin216x+2 víi x � 0;   Chuyªn ®Ò ph¬ng trinh lîng gi¸c 2 � �  k 2 � 4/ 8cos4xcos22x+ 1  cos 3x +1=0 x  � � � 3 4 ........ 5/  sin x  cos x x0 8/( cos2x-cos4x)2=6+2sin3x Chuyªn ®Ò ph¬ng trinh lîng gi¸c 6/ 5-4sin2x-8cos2x/2 =3k t×m k �Z* ®Ó hÖ cã nghiÖm x   k 2 9/   1 1  cos x  1  cos x cos 2 x  sin 4 x 2 2 7/ 1- x =cosx 2  x  �  k 2 4 5
- Xem thêm -

Tài liệu liên quan