M u n d a r I j a
Download 1.09 Mb. Pdf ko'rish
|
abiturshtabalgebra
· cos(x − 60
0 ) = b ⇐⇒ cos(x − 60 0 ) = b tenglamaga ega bo’lamiz. Bu tenglama faqat b ∈ [−1; 1] larda yechimga ega. Javob: −1 ≤ b ≤ 1 (B). 34. (98-9-25) Tenglama k ning qanday qiymatlarida yechimga ega? sin(60 0 + x) − sin(60 0 − x) = k A) k ∈ (−1; 1) B) k ∈ [−1; 1] C) k ≤ 1 D) k ≤ −1 35. (02-7-18) Tenglamani yeching. sin 5x + sin 3x + sin x = 0 A) πn 3 ; ± π 3 + πn, n ∈ Z B) nπ 3 ; π 2 + πn 2 , n ∈ Z C) π 2 + πn 2 , n ∈ Z D) πn 3 , n ∈ Z Yechish: Sinuslar yig’indisi sin 5x + sin x ni ko’- paytmaga almashtirib, berilgan tenglamaga teng kuchli bo’lgan 2 sin 3x · cos 2x + sin 3x = 0 yoki sin 3x(2·cos 2x+1) = 0 tenglamaga ega bo’lamiz. Bu tenglamaning yechimlari sin 3x = 0 va cos 2x = − 1 2 tenglama yechimlari birlashmasidan iborat. Bu tenglamalar eng sodda trigonometrik tenglamalar bo’lib, ularning yechimlari 2 va 5-formulalar yor- damida topiladi: x = πn 3 , x = ± π 3 + πn, n ∈ Z. Javob: πn 3 ; ± π 3 + πn, n ∈ Z. (A). 36. (00-10-57) Ushbu sin 2x + sin 4x = 0 tenglama [0; 2π] oraliqda nechta ildizga ega? A) 0 B) 7 C) 4 D) 9 37. (02-1-61) Tenglamani yeching. sin 6x + sin 2x = sin 4x A) πn 4 , n ∈ Z B) π 3 + 2πn, n ∈ Z C) − π 3 + πn, n ∈ Z D) πn 4 , ± π 6 + πn, n ∈ Z 38. (03-6-63) Qanday eng kichik o’tkir burchak sin(2x + 45 0 ) = cos(30 0 − x) tenglamani qanoatlantiradi? A) 25 0 B) 5 0 C) 45 0 D) 15 0 39. (98-1-59) Tenglama [0; π] kesmada nechta ildizga ega?. cos x · cos 4x − cos 5x = 0 A) 1 B) 2 C) 4 D) 5 Yechish: cos 5x = cos(x + 4x) ga qo’shish for- mulasini (13.2.3-ning 3-ga qarang) qo’llab, keyin 161 o’xshash hadlarni ixchamlab berilgan tenglamaga teng kuchli bo’lgan sin x · sin 4x = 0 tenglamaga ega bo’lamiz. Bu tenglamadan sin x = 0 yoki sin 4x = 0 ni olamiz. sin x = 0 tenglama [0; π] kesmada ikkita 0 va π ildizlarda ega, sin 4x = 0 tenglama esa [0; π] kesmada 5 ta 0, π 4 , 2π 4 , 3π 4 va 4π 4 = π ildizlarda ega. 1-tenglamaning ildi- zlari 0 va π lar, 2-tenglamaning yechimlari ichida uchraganligi uchun ular bir marta sanaladi. De- mak, berilgan tenglama [0; π] kesmada 5 ta ildizga ega ekan. Javob: 5 (D). 40. (98-8-59) Tenglama [0; 2π] oraliqda nechta ildizga ega? cos x · cos 2x = cos 3x A) 5 B) 4 C) 3 D) 2 41. Tenglamaning I va II chorakdagi ildizlari yig’in- disini toping? sin(3x − 45 0 ) = 0 A) 135 0 B) 150 0 C) 210 0 D) 225 0 42. (02-1-19) Tenglamaning [0; π 2 ] kesmadagi ildizlari yig’indisini toping? cos 4x · cos 5x = cos 6x · cos 7x A) 41π 22 B) 31π 22 C) 30π 11 D) 43π 22 43. (02-10-60) Tenglamani yeching. cos à 3π + x 3 ! · cos à 9π + 2x 6 ! = 1 4 A) (−1) n+1 π 3 + 2πn; n ∈ Z B) (−1) n+1 π 6 + πn; n ∈ Z C) (−1) n π 3 + 2πn; n ∈ Z D) (−1) n π 6 + 3πn 2 ; n ∈ Z 4. Bir xil ismga keltiriladigan tenglamalar 44. (97-1-46) Tenglamani yeching. 2 cos 2 (x − π) + 3 sin(π + x) = 0 A) π 2 + πn, n ∈ Z B) (−1) n π 6 + πn, n ∈ Z C) ± π 3 + 2πn, n ∈ Z D) ± π 6 + 2πn, n ∈ Z Yechish: Keltirish formulalari cos(x − π) = − cos x; sin(π + x) = − sin x ni qo’llab, berilgan tenglamani 2 cos 2 x − 3 sin x = 0 shaklda yozib olamiz. Bu tenglamani xil ismga keltirish uchun cos 2 x = 1 − sin 2 x ayniyatdan foydalanamiz, natijada 2 sin 2 x + 3 sin x − 2 = 0 tenglamani olamiz. Agar sin x = y belgilash ol- sak oxirga tenglama 2y 2 + 3y − 2 = 0 kvadrat tenglamaga keladi. Bu kvadrat tenglamaning ildi- zlari y 1 = −2 va y 2 = 0, 5 lardir. sin x = y 1 = −2 tenglama ildizga ega emas, chunki | − 2| > 1. sin x = y 2 = 0, 5 tenglamaning ildizlari x = (−1) n π 6 + πn, n ∈ Z ko’rinishga ega. Javob: (B). 45. (97-11-45) Tenglamani yeching. 2 sin 2 (π − x) + 5 sin(1, 5π + x) = 2 A) πn, n ∈ Z B) π 2 + πn, n ∈ Z C) π 2 + 2πn, n ∈ Z D) (−1) n · π 6 + πn, n ∈ Z 46. (97-1-50) Tenglamaning (0 0 ; 90 0 ] oraliqdagi ildizini toping. 2 sin 2 x − √ 3 sin 2x = 0 A) 30 0 B) 45 0 C) 60 0 D) 90 0 47. (00-3-52) Tenglamaning [0; 2π] kesmadagi eng katta va eng kichik ildizlari ayirmasini toping? cos 2 x − 1 2 sin 2x = 0 A) π 2 B) 3π 4 C) π D) 5π 4 48. (02-3-79) Tenglama [−2π; π] kesmada nechta il- dizga ega?. tgx + 1 tgx = 2 A) 3 B) 5 C) 4 D) 6 49. (00-5-41) Tenglamani yeching. cos 2x − 5 sin x − 3 = 0 A) (−1) n π 6 + πn, n ∈ Z B) (−1) n+1 π 6 + πn, n ∈ Z C) (−1) n π 6 + 2πn, n ∈ Z D) (−1) n+1 π 6 + 2πn, n ∈ Z Yechish: cos 2x = 1 − 2 sin 2 x ayniyatdan foy- dalanib, berilgan tenglamani 2 sin 2 x + 5 sin x + 2 = 0 shaklda yozib olamiz. Bu tenglamada sin x = y belgilash olib, uni 2y 2 + 5y + 2 = 0 kvadrat tenglamaga keltiramiz. Bu kvadrat tenglamaning ildizlari y 1 = −2 va y 2 = −0, 5 lardir. sin x = 162 y 1 = −2 tenglama ildizga ega emas, chunki | − 2| > 1. sin x = y 2 = −0, 5 tenglamaning ildizlari x = (−1) n+1 π 6 + πn, n ∈ Z ko’rinishga ega. Javob: (B). 50. (02-11-43) Tenglamaning (−90 0 ; 180 0 ) intervalga tegishli ildizlari yig’indisini toping. 3 sin 2 2x + 7 cos 2x − 3 = 0 A) 90 0 B) 105 0 C) 180 0 D) 135 0 51. (02-11-44) Tenglamaning [−4π; 4π] kesmaga te- gishli ildizlari nechta? cos 2x + 5 cos x = 6 A) 4 B) 5 C) 6 D) 8 52. (02-12-40) Tenglamaning [0; 2π] kesmadagi ildiz- lari yig’indisini hisoblang. cos 2x − 2 sin 2 x = 0 A) 3, 5π B) 3 1 6 π C) 4π D) 4 1 6 π 53. (03-4-25) Tenglama ildizlari yig’indisini toping. 1 − sin x − cos 2x = 0 (x ∈ [0; 2π]) A) 3, 5π B) 4, 2π C) 4π D) 3, 8π 5. Darajani pasaytirish usuli yordamida yechiladigan tenglamalar Darajani pasaytirish formulalari quyidagilardir: • sin 2 x = 1 − cos 2x 2 ⇐⇒ 1−cos 2x = 2 sin 2 x. • cos 2 x = 1 + cos 2x 2 ⇐⇒ 1+cos 2x = 2 cos 2 x. 54. (98-2-26) Tenglamani yeching. 2 cos 2 x − 1 = − 1 2 A) (−1) k π 6 + π 2 k; k ∈ Z B) (−1) k+1 π 6 + πk, k ∈ Z C) ± π 6 + πk, k ∈ Z D) ± π 3 + πk, k ∈ Z Yechish: Darajani pasaytirish formulasining 2- dan foydalanib, berilgan tenglamani cos 2x = −0, 5 shaklda yozib olamiz. Bu tenglama eng sodda trigonometrik tenglama bo’lib, uning ildizlari x = ± π 3 + πn, n ∈ Z ko’rinishga ega. Javob: (D). 55. (98-6-50) Tenglamani yeching. 4 cos 2 2x − 1 = cos 4x A) π 4 + πn 2 , n ∈ Z B) πn 2 , n ∈ Z C) π 6 + πn 2 , n ∈ Z D) π 3 + πn 2 , n ∈ Z 56. (96-9-50) Ushbu 4 sin x 2 − cos x + 1 = 0 tenglamaning [0; 2π] kesmada nechta ildizi bor? A) 0 B) 2 C) 3 D) 1 57. (96-12-97) Ushbu sin x 2 + cos x − 1 = 0 tenglama [0; 2π] oraliqda nechta yechimga ega? A) 3 B) 4 C) 0 D) 2 58. (96-13-43) Tenglamaning [0; 2π] kesmada nechta ildizi bor? 4 cos x 2 + cos x + 1 = 0 A) 1 B) 2 C) 0 D) 3 59. (98-11-99) Tenglamani yeching. 2 cos 2 x 2 = 1 + cos x + cos 2x A) π 4 + πk 2 , k ∈ Z B) π 4 + πk, k ∈ Z C) πk 2 , k ∈ Z D) πk, k ∈ Z 60. (01-1-48) Tenglamani yeching. 4 sin 2 x(1 + cos 2x) = 1 − cos 2x A) πn, n ∈ Z B) πn; ± π 3 + πn, n ∈ Z C) ± π 3 + πn, n ∈ Z D) πn; ± π 3 + 2πn, n ∈ Z Yechish: Darajani pasaytirish formulasining 1- dan foydalanib, berilgan tenglamani 2(1 − cos 2x)(1 + cos 2x) = 1 − cos 2x shaklda yozib olamiz. Bu ifodada 1−cos 2x umu- miy ko’paytuvchini qavs oldiga chiqarib (1 − cos 2x)(1 + 2 cos 2x) = 0 tenglamani olamiz. Bu yerdan cos 2x = 1 yoki cos 2x = −0, 5 tenglamaga kelamiz. Bular eng sodda trigonometrik tenglamalar bo’lib, ularning ildizlari x = πn, n ∈ Z; x = ± π 3 + πn, n ∈ Z ko’rinishga ega. Javob: (B). 163 61. (99-10-34) Tenglamani yeching. (1 + cos x)tg x 2 = 0 A) πk, k ∈ Z B) π + 2πk, k ∈ Z C) 2πk, k ∈ Z D) π + πk, k ∈ Z 62. (01-2-81) Ushbu 7 cos 2x − 6 = cos 4x tenglamaning [0; 628] kesmaga tegishli ildizlari yig’indisini toping. A) 200π B) 199π C) 20100π D) 19900π 63. (02-6-44) Tenglama [0; 2π] kesmada nechta ildizga ega? 3 sin 2x − 2 cos 2x = 2 A) 5 B) 1 C) 2 D) 4 64. (03-10-41) Tenglamani yeching. sin 2 x + sin 2 4x = sin 2 2x + sin 2 3x A) πn 2 , n ∈ Z B) π 5 + 2πn 5 , n ∈ Z C) π 10 + 2πn 5 , n ∈ Z D) π 10 + πn 5 ; πn 2 , n ∈ Z 6. Quyidagi tenglamalarni yechishda uning aniqlanish sohasiga e’tibor bering 65. (98-1-56) Tenglamani yeching. sin 2x tgx − 1 = 0 A) πk 2 , k ∈ Z B) π 2 + πk, k ∈ Z C) 2πk, k ∈ Z D) πk, k ∈ Z Yechish: Berilgan tenglama tgx − 1 6= 0, cos x 6= 0 shartda aniqlangan. Kasr nolga aylanishi uchun uning surati, ya’ni sin 2x = 0 bo’lishi kerak. Bu tenglamani sin 2α = 2 sin α cos α ekanidan foy- dalanib 2 sin x cos x = 0 ko’rinishda yozamiz. Bu yerdan cos x 6= 0 ni e’tiborga olib, sin x = 0 tenglamani, bundan esa x = πk, k ∈ Z ekanini hosil qilamiz. Bu nuqtalarda tgx − 1 6= 0 shart ham bajariladi. Javob: πk, k ∈ Z (D). 66. (97-7-59) Tenglama [0; 4π] oraliqda nechta ildizga ega? sin 2 x + sin x cos x = 0 A) 5 B) 4 C) 7 D) 2 67. (97-12-65) Tenglama [−2π; 2π] oraliqda nechta yechimga ega? cos 2 x − cos x sin x = 0 A) 6 B) 4 C) 3 D) 2 68. (98-9-26) Tenglamani yeching. 1 cos 2 x = 2tg 2 x A) ± π 4 + 2πk, k ∈ Z B) ± π 4 + πk, k ∈ Z C) ± π 3 + πk, k ∈ Z D) ± π 3 + 2πk, k ∈ Z 69. (99-1-44) ctg(x + 1) · tg(2x − 3) = 1 tenglamaning [π; 2π] oraliqdagi yechimini toping. A) 4 B) 2 C) 3 D) 5 70. (00-4-47) Tenglamaning [π; 3π] kesmadagi ildi- zlari yig’indisini toping. √ 1 − cos x = sin x A) 2π B) 5π C) 6π D) 4, 5π 71. (98-10-105) Tenglamaning [0; 2π] kesmada nechta ildizi bor? 1 + cos x sin x = cos x 2 A) 0 B) 1 C) 2 D) 3 72. (01-2-32) Tenglamani yeching. cos 3x sin 3x − 2 sin x = tgx A) π 4 + πn, n ∈ Z B) π 4 + 2πn, n ∈ Z C) π 4 + π 2 n, n ∈ Z D) π 3 + π 2 n, n ∈ Z 73. (01-6-30) Tenglamaning [0; 4π] kesmadagi ildiz- lari yig’indisini toping. tg 2 x − 2 cos x + 1 = 0 A) 7π B) 7 2 3 π C) 8π D) 7 1 3 π Yechish: Berilgan tenglama cos x 6= 0 shartda aniqlangan. 1 + tg 2 x = 1 cos 2 x ayniyatdan foy- dalanib berilgan tenglamani 1 cos 2 x − 2 cos x = 0 ⇐⇒ 1 − 2 cos x cos x = 0 ko’rinishda yozamiz. Bu yerdan cos x = 0, 5 ni, bundan esa x = ± π 3 + 2πk, k ∈ Z ekanini hosil qilamiz. Bu yechimlardan 4 tasi π 3 ; 2π ± π 3 ; 4π − π 3 lar [0; 4π] kesmada yotadi. Ularning yig’indisi 8π. Javob: 8π (C). 164 74. (01-10-37) Tenglamaning [− π 2 ; π 2 ] kesmada nechta ildizi bor? cos 4x + 10tgx 1 + tg 2 x = 3 A) 0 B) 1 C) 2 D) 3 75. (01-11-21) Tenglamani yeching. tgx tg3x = −1 A) π 2 k, k ∈ Z B) πk, k ∈ Z C) π 4 + π 2 k, k ∈ Z D) π 4 + πk, k ∈ Z 76. (98-3-58) Tenglamaning [0; 4π] kesmada nechta ildizi bor? cos 2x √ 2 2 + sin x = 0 A) 8 B) 6 C) 4 D) 2 77. (03-7-39) Tenglamani yeching. q cos 2x + √ 3 sin x = −2 cos x A) 2π 3 + 2kπ, k ∈ Z B) π 3 + 2kπ, k ∈ Z C) (−1) k π 3 + πk, k ∈ Z D) (−1) k 2π 3 + 2πk, k ∈ Z 7. Turli tenglamalar 78. (98-5-50) Tenglamani yeching. 4 cos 2 x+2 cos x = 1 A) πn; π 2 + 2πn, n ∈ Z B) π 2 + πn, n ∈ Z C) πn; − π 2 + 2πn, n ∈ Z D) 2πn, n ∈ Z Yechish: Berilgan tenglamani 4 cos 2 x+2 cos x = 4 0 shaklda yozamiz. Bu tenglama cos 2 x + 2 cos x = 0 ⇐⇒ (cos x + 2) cos x = 0 tenglamaga teng kuchli. cos x + 2 6= 0 bo’lganligi uchun cos x = 0 bo’lib, uning yechimlari x = π 2 + πn, n ∈ Z. Javob: π 2 + πn, n ∈ Z (B). 79. (99-7-48) Tenglamani yeching. 5 · 5 sin 2 x+cos 2x = 1 25 A) ∅ B) πn, n ∈ Z C) π 2 + 2πn, n ∈ Z D) 2πn, n ∈ Z 80. (97-3-58) Tenglamani yeching. 2 1−log 2 sin x = 4 A) π 6 + 2πn; n ∈ Z B) (−1) n π 6 + πn, n ∈ Z C) (−1) n π 3 + πn, n ∈ Z D) π 4 + 2πn, n ∈ Z 81. (97-7-58) Tenglamani yeching. 3 1+log 3 ctgx = √ 3 A) π 6 + πn; n ∈ Z B) π 3 + πn, n ∈ Z C) π 3 + 2πn, n ∈ Z D) π 4 + πn, n ∈ Z 82. (99-2-37) a ning qanday qiymatlarida log a sin x = 1 tenglama yechimga ega? A) a ∈ [−1; 1] B) a ∈ (−1; 1) C) a ∈ (0; 1] D) a ∈ (0; 1) 83. (02-9-36) Tenglamani yeching. 9 cos x + 2 · 3 cos x = 15 A) πn, n ∈ Z B) 2πn, n ∈ Z C) π 3 + 2πn, n ∈ Z D) π 2 + πn, n ∈ Z 84. (03-5-41) Tenglamani yeching. 8 sin 2 x − 2 cos 2 x = 0 A) ± π 6 + πn, n ∈ Z B) π 6 + πn, n ∈ Z C) − π 6 + πn, n ∈ Z D) π 4 + πn, n ∈ Z 85. (03-12-61) a parametrning qanday qiymatlarida sin 6 x + cos 6 x = a tenglama yechimga ega? A) [0; 1] B) [0, 5; 1] C) [0, 25; 0, 5] D) [0, 25; 1] 13.5 Trigonometrik tengsizliklar sin x ≥ a, cos x ≥ a, tgx ≥ a, ctgx ≥ a tengsizliklar sodda trigonometrik tengsizliklar deyiladi. Bu yerda tengsizlik > yoki ≤ yoki < belgilaridan ixtiyoriy biri bo’lishi mumkin. Biz asosan ayniy almashtirishlar yor- damida sodda trigonometrik tengsizliklarga yoki shu tipdagi tengsizliklar sistemasiga keladigan tengsizlik- larni yechish usullarini beramiz. Sodda trigonometrik tengsizliklarning yechimlarini keltiramiz. 1. sin Download 1.09 Mb. Do'stlaringiz bilan baham: |
Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling
ma'muriyatiga murojaat qiling