1.6-bo'lim : Trig tenglamalarni kalkulyator yordamida yechish, II qism
5. \(7{\cos ^2}\left( {3x} \right) - \cos \left( {3x} \right) = 0\) ning barcha yechimlarini toping. Ishingizda kamida 4 kasrdan foydalaning.
Yechim
E'tibor bering, biz har bir atama uchun kosinusni hisobga olamiz,
\[\cos \left( {3x} \right)\left( {7\cos \left( {3x} \right) - 1} \right) = 0\]Endi bizda nolga teng bo'lgan ikkita omil mahsuloti bor va shuning uchun asosiy algebraik xususiyatlar bo'yicha biz ega bo'lishimiz kerakligini bilamiz,
\[\cos \left( {3x} \right) = 0\hspace{0.25in}{\rm{OR}}\hspace{0.25in}7\cos \left( {3x} \right) - 1 = 0\]2-qadam
Bu tenglamalarning har biri oldingi bobda yechilgan tenglamalarga o'xshaydi. Shuning uchun, biz har biri uchun yechim jarayonini eslab qolishingiz mumkin deb o'ylaymiz va biz unchalik ko'p tafsilotlarni kiritmaymiz. Agar jarayonga ishonchingiz komil bo'lmasa, oldingi bo'limga qaytib, ushbu muammoni hal qilishda davom etishdan oldin ba'zi muammolar bilan shug'ullanishingiz kerak.
Biz boshlaymiz,
\[\cos \left( {3x} \right) = 0\]Birlik doiradan biz ko'rishimiz kerakki,
\[3x = \frac{\pi }{2} + 2\pi n\hspace{0.25in}{\rm{OR}}\hspace{0.25in}3x = \frac{{3\pi }}{2} + 2\pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]E'tibor bering, biz buni yanada kamaytirishimiz mumkin,
\[3x = \frac{\pi }{2} + \pi n\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]Nihoyat, bu tenglamaning yechimlari:
\[x = \frac{\pi }{6} + \frac{{\pi n}}{3}\hspace{0.25in}\,\,n = 0, \pm 1, \pm 2, \ldots \]Ikkinchi tenglama biroz ko'proq (lekin ko'p emas) ish talab qiladi. Birinchidan, kosinusni ajratish,
\[\cos \left( {3x} \right) = \frac{1}{7}\]Kalkulyatorimiz yordamida biz quyidagilarni olamiz:
\[3x = {\cos ^{ - 1}}\left( {\frac{1}{7}} \right) = 1.4274\]Birlik doirasiga tez nazar tashlasak, \(\left[ {0,2\pi } \right]\) diapazonidagi ikkinchi burchak \(2\pi - 1.4274 = 4.8558\) bo'lishini bilamiz.
Nihoyat, bu tenglamaning yechimlari:
\[\begin{align*}3x & = 1.4274 + 2\pi n & \hspace{0.25in}{\rm{OR}}\hspace{0.25in} 3x & = 4.8558 + 2\pi n & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \\x & = 0.4758 + \frac{{2\pi n}}{3} & \hspace{0.25in}{\rm{OR}}\hspace{0.25in} x & = 1.6186 + \frac{{2\pi n}}{3} & \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots \end{align*}\]Bularning barchasini birlashtirib, quyidagi echimlar to'plamini beradi.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{\pi }{6} + \frac{{\pi n}}{3},\hspace{0.25in} x = 0.4758 + \frac{{2\pi n}}{3},\hspace{0.25in} {\rm{OR }}\hspace{0.25in} x = 1.6186 + \frac{{2\pi n}}{3}\hspace{0.25in}n = 0, \pm 1, \pm 2, \ldots }}\]Agar interval berilgan bo'lsa, biz \(n\) qiymatlarini o'rnatishni davom ettiramiz va bu oraliqda qaysi echimlar to'g'ri kelishini aniqlashimiz mumkin. Bizga oraliq berilmagani uchun bu biz imkon qadar uzoqdir.
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.