TrigEquations - 1-masala yechimi | MathModel.uz

1.4-bo'lim : Trig tenglamalarini yechish

1. Kalkulyatordan foydalanmasdan $4\sin \left( {3t} \right) = 2$ ning barcha yechimlarini toping.

Yechim

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

\[\sin \left( {3t} \right) = \frac{1}{2}\]

2-qadam

Biz bu muammoda sinus bilan shug‘ullanayotganimiz va $y$ o‘qi birlik aylanada sinusni ifodalashini bilamiz, biz ${\textstyle{1 \over 2}}$ ning $y$ koordinatasiga ega bo‘lgan burchaklarni qidiramiz. Bu shuni anglatadiki, biz birinchi kvadrantda burchakka va ikkinchi kvadrantda burchakka ega bo'lamiz (buni topish uchun birinchi kvadrantdagi burchakdan 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 the angle $\frac{{5\pi }}{6}$ in the second quadrant, and a horizontal dashed line joins the point where $\frac{\pi }{6}$ meets the circle to the point where $\frac{{5\pi }}{6}$ meets it, showing that the two angles have the same y-coordinate of $\frac{1}{2}$ and so the same sine value.$

Shubhasiz, birinchi kvadrantdagi burchak $\frac{\pi }{6}$ va ba'zi bir asosiy simmetriyaga ko'ra, ikkinchi burchak uchun terminal chizig'i yuqorida ko'rsatilgandek manfiy $x$ o'qi bilan $\frac{\pi }{6}$ burchak hosil qilishi kerakligini ko'rishimiz mumkin va shunday bo'ladi: $\pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$.

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,

\[3t = \frac{\pi }{6} + 2\pi n \hspace{0.5in}{\rm{OR }}\hspace{0.5in}3t = \frac{{5\pi }}{6} + 2\pi n \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots \]

Nihoyat, tenglamaning barcha yechimlarini olish uchun ikkala tomonni 3 ga bo'lish kifoya.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{\pi }{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} {\rm{OR }} \hspace{0.5in} t = \frac{{5\pi }}{{18}} + \frac{{2\pi n}}{3} \hspace{0.5in} n = 0, \pm 1, \pm 2, \ldots }}\]