TrigEquations - 3-masala yechimi | MathModel.uz

1.4-bo'lim : Trig tenglamalarini yechish

3. Kalkulyatordan foydalanmasdan $\displaystyle 2\cos \left( {\frac{x}{3}} \right) + \sqrt 2 = 0$ ning barcha yechimlarini toping.

Yechim

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

\[\cos \left( {\frac{x}{3}} \right) = - \frac{{\sqrt 2 }}{2}\]

2-qadam

Biz bu muammoda kosinus bilan shug‘ullanayotganimiz va $x$ o‘qi birlik aylanada kosinus ekanligini bilamiz, biz $ - \frac{{\sqrt 2 }}{2}$ ning $x$ koordinatasiga ega bo‘lgan burchaklarni qidiramiz. Bu shuni anglatadiki, biz ikkinchi va uchinchi kvadrantda burchaklarga ega bo'lamiz.

Salbiy qiymat tufayli biz birinchi kvadrantda mos keladigan burchakni topa olmaymiz va ikkinchi burchakni topish uchun undan foydalanamiz. Biroq, bizga kerak bo'lgan ikkita burchakni topish uchun birinchi kvadrantdagi burchaklardan foydalanishimiz mumkin. Mana bu holat uchun birlik doirasi.

$A unit circle with the first quadrant angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ drawn in and labeled with their coordinates. Also drawn in is a single line through the origin that meets the circle at $\frac{{3\pi }}{4}$ in the second quadrant and at $\frac{{5\pi }}{4}$ in the third quadrant. Dashed lines show that both of these angles have an x-coordinate of $ - \frac{{\sqrt 2 }}{2}$ and so both give the same cosine value.$

Agar bizda salbiy qiymat bo'lmasa, burchak $\frac{\pi }{4}$ bo'ladi. Endi, birlik aylanasidagi simmetriyaga asoslanib, ikkala burchak uchun terminal chizig'i manfiy $x$ o'qi bilan $\frac{\pi }{4}$ burchagini hosil qiladi. Keyin ikkinchi kvadrantdagi burchak $\pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$, uchinchi kvadrantdagi burchak esa $\pi + \frac{\pi }{4} = \frac{{5\pi }}{4}$ bo'ladi.

3-qadam

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

Bu shuni anglatadiki, bizda bo'lishi kerak,

\[\frac{x}{3} = \frac{{3\pi }}{4} + 2\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} \frac{x}{3} = \frac{{5\pi }}{4} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Nihoyat, tenglamaning barcha yechimlarini olish uchun ikkala tomonni 3 ga ko'paytirish kifoya.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{9\pi }}{4} + 6\pi n \hspace{0.5in} {\rm{OR }} \hspace{0.5in} x = \frac{{15\pi }}{4} + 6\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots }}\]