Asosiy tarkibga o'tish

1.5-bo'lim : Trig tenglamalarni kalkulyator yordamida yechish, I qism

1. \(7\cos \left( {4x} \right) + 11 = 10\) ning barcha yechimlarini toping. Ishingizda kamida 4 kasrdan foydalaning.

Yechim

Kosinusni (bir koeffitsient bilan) tenglamaning bir tomonida ajratib olish,

\[\cos \left( {4x} \right) = - \frac{1}{7}\]
2-qadam

Birinchidan, kalkulyatorimiz yordamida biz buni ko'rishimiz mumkin,

\[4x = {\cos ^{ - 1}}\left( { - \frac{1}{7}} \right) = 1.7141\]

Endi biz bu muammoda kosinus bilan shug‘ullanmoqdamiz va biz \(x\) o‘qi birlik aylanada kosinusni ifodalashini bilamiz va shuning uchun biz \( - \frac{1}{7}\) ning \(x\) koordinatasiga ega bo‘lgan burchaklarni qidiramiz. Bu shuni anglatadiki, biz ikkinchi (bu bizning kalkulyatorimiz bergan) va uchinchi kvadrantda burchaklarga ega bo'lamiz. Mana bu holat uchun birlik doirasi.

A unit circle with the axes labeled with the angles 0 and \(2\pi = 6.2832\), \(\frac{\pi }{2} = 1.5708\), \(\pi = 3.1416\) and \(\frac{{3\pi }}{2} = 4.7124\). Two lines from the center meet the circle just to the left of the \(y\)-axis, one in the second quadrant labeled 1.7141 and one in the third quadrant labeled 4.5691. A dashed vertical line connects the two points, showing they have the same \(x\)-coordinate of \( - \frac{1}{7}\).

Birlik doirasining simmetriyasidan biz ikkinchi burchak uchun -1,7141 yoki \(2\pi - 1.7141 = 4.5691\) dan foydalanishimiz mumkinligini ko'rishimiz mumkin. Har biri bir xil echimlar to'plamini beradi. Biroq, minus belgilarini yo'qotish oson bo'lganligi sababli, biz ikkinchi yechim uchun ijobiy burchakdan foydalanamiz.

3-qadam

Ushbu bo'lim uchun eslatmalar muhokamasidan shuni bilamizki, biz ushbu ikki burchakka ega bo'lgach, ularning har biriga "\( + \,2\pi n\) for \(n = 0, \pm 1, \pm 2, \ldots \)" ni qo'shish orqali barcha mumkin bo'lgan burchaklarni olishimiz mumkin.

Bu shuni anglatadiki, bizda bo'lishi kerak,

\[4x = 1.7141 + 2\pi n\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}4x = 4.5691 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots \]

Nihoyat, tenglamaning barcha yechimlarini olish uchun ikkala tomonni 4 ga bo'lish kifoya.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0.4285 + \frac{{\pi n}}{2}\hspace{0.25in}{\mbox{OR }}\hspace{0.25in}x = 1.1423 + \frac{{\pi n}}{2}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}\]

E'tibor bering, bu erda ishlatgan o'nli kasrlar miqdoriga qarab, javoblaringiz yaxlitlash xatosi tufayli ulardan biroz farq qilishi mumkin. Har qanday farqlar ozgina bo'lishi kerak va faqat 4 atrofida paydo bo'lishi kerakth o'nlik kasr yoki shunga o'xshash.