Trigonometrik tenglamalar

Ushbu darsda burchak (yoy) trigonometrik funksiyasi ostida noma’lum qatnashgan tenglamalar haqida so‘z yuritamiz. Trigonometrik funksiyalar davriy bo‘lgani uchun bunday tenglamalarning yechimi cheksiz ko‘p bo‘ladi — biz ularni umumiy formula ko‘rinishida yozamiz.

§1. Nima uchun cheksiz yechim?

Savol. $\sin x = \tfrac{1}{2}$ tenglamada nechta yechim bor? Sinus $2\pi$ davr bilan takrorlangani uchun har bir yechimga $2\pi$ ni qo‘shsak, yana yechim chiqadi — demak ular cheksiz ko‘p.

Ta’rif. Noma’lum $x$ biror trigonometrik funksiya ($\sin, \cos, \tan, \cot$) argumentida qatnashgan tenglama trigonometrik tenglama deyiladi.

§2. Eng sodda tenglamalar formulalari

Qoida. $|a|\le 1$ bo‘lganda ($n \in \mathbb{Z}$):
$\sin x = a \Rightarrow x = (-1)^n \arcsin a + \pi n$;
$\cos x = a \Rightarrow x = \pm \arccos a + 2\pi n$;
$\tan x = a \Rightarrow x = \arctan a + \pi n$ (har qanday $a$ uchun).

Qoida. Muhim xususiy hollar:
$\sin x = 0 \Rightarrow x = \pi n$; $\ \sin x = 1 \Rightarrow x = \tfrac{\pi}{2} + 2\pi n$; $\ \sin x = -1 \Rightarrow x = -\tfrac{\pi}{2}+2\pi n$.
$\cos x = 0 \Rightarrow x = \tfrac{\pi}{2} + \pi n$; $\ \cos x = 1 \Rightarrow x = 2\pi n$; $\ \cos x = -1 \Rightarrow x = \pi + 2\pi n$.

Eslatma. Agar $|a|>1$ bo‘lsa, $\sin x = a$ yoki $\cos x = a$ tenglamaning yechimi yo‘q, chunki bu funksiyalar $[-1;1]$ oralig‘idan chiqmaydi.

Misol. $\sin x = \tfrac{1}{2}$ tenglamani yeching.
Yechish. $\arcsin \tfrac{1}{2} = \tfrac{\pi}{6}$, formulaga qo‘yamiz:

Javob. $x = (-1)^n \tfrac{\pi}{6} + \pi n$.

Misol. $\cos x = -\tfrac{\sqrt{2}}{2}$ tenglamani yeching.
Yechish. $\arccos\!\left(-\tfrac{\sqrt{2}}{2}\right) = \tfrac{3\pi}{4}$:

Javob. $x = \pm \tfrac{3\pi}{4} + 2\pi n$.

Misol. $\tan x = 1$ tenglamani yeching.
Yechish. $\arctan 1 = \tfrac{\pi}{4}$:

Javob. $x = \tfrac{\pi}{4} + \pi n$.

§3. Kvadrat tenglamaga keltirish

Misol. $2\sin^2 x - 3\sin x + 1 = 0$ tenglamani yeching.
Yechish. $t = \sin x$, bunda $-1 \le t \le 1$:

$\sin x = \tfrac{1}{2} \Rightarrow x = (-1)^n \tfrac{\pi}{6} + \pi n$; $\ \sin x = 1 \Rightarrow x = \tfrac{\pi}{2} + 2\pi n$.

Javob. $x = (-1)^n \tfrac{\pi}{6} + \pi n$ yoki $x = \tfrac{\pi}{2} + 2\pi n$.

Misol. $\cos 2x + \cos x = 0$ tenglamani yeching.
Yechish. $\cos 2x = 2\cos^2 x - 1$ formulasidan foydalanamiz. $t=\cos x$:

$t=\tfrac{1}{2} \Rightarrow \cos x = \tfrac{1}{2} \Rightarrow x = \pm\tfrac{\pi}{3} + 2\pi n$; $\ t=-1 \Rightarrow \cos x = -1 \Rightarrow x = \pi + 2\pi n$.

Javob. $x = \pm \tfrac{\pi}{3} + 2\pi n$ yoki $x = \pi + 2\pi n$.

Eslatma. Yangi o‘zgaruvchi $t=\sin x$ yoki $t=\cos x$ da $|t|\le 1$ shartini tekshiring — bu oraliqdan tashqaridagi ildizlar mos kelmaydi.

§ Lug‘at

trigonometrik tenglama — trigonometric equation
davr (davriylik) — period
yoy funksiyalari — inverse trig functions
xususiy hol — special case
qo‘sh burchak formulasi — double-angle formula
yechimlar to‘plami — solution set