Trigonometrik tengsizliklar

Ushbu darsda trigonometrik funksiya qatnashgan tengsizliklar haqida so‘z yuritamiz. Bunday tengsizliklarni yechishning eng ishonchli yo‘li — birlik aylana yoki funksiya grafigidan foydalanish. Chunki davriylik tufayli yechim cheksiz ko‘p oraliqdan iborat bo‘ladi.

§1. Birlik aylana yordamida fikrlash

Ta’rif. Trigonometrik funksiya argumentida noma’lum qatnashgan tengsizlik trigonometrik tengsizlik deyiladi. Masalan $\sin x > \tfrac{1}{2}$.

Savol. $\sin x$ nima uchun aylana bilan bog‘liq? Birlik aylanada burchak $x$ ga mos nuqtaning ordinatasi aynan $\sin x$ ga teng. Demak $\sin x > \tfrac{1}{2}$ — bu ordinatasi $\tfrac{1}{2}$ dan katta nuqtalarning yoyi.

Qoida. Yechish tartibi: 1) tegishli sodda tenglamaning ildizlarini toping; 2) birlik aylanada (yoki grafikda) tengsizlik bajariladigan yoyni belgilang; 3) davrni ($2\pi$ yoki $\pi$) qo‘shib, umumiy javobni yozing.

§2. Sinus va kosinus tengsizliklari

Misol. $\sin x > \tfrac{1}{2}$ tengsizlikni yeching.
Yechish. $\sin x = \tfrac{1}{2}$ ning bir davrdagi ildizlari $x=\tfrac{\pi}{6}$ va $x=\tfrac{5\pi}{6}$. Aylanada ordinata $\tfrac{1}{2}$ dan katta yoy shu ikki nuqta orasida (yuqorida) yotadi:

Javob. $x \in \left(\tfrac{\pi}{6}+2\pi n;\ \tfrac{5\pi}{6}+2\pi n\right)$.

Misol. $\cos x \le -\tfrac{\sqrt{2}}{2}$ tengsizlikni yeching.
Yechish. $\cos x = -\tfrac{\sqrt{2}}{2}$ ning ildizlari $x = \tfrac{3\pi}{4}$ va $x = \tfrac{5\pi}{4}$. Kosinus — nuqtaning abssissasi; u $-\tfrac{\sqrt{2}}{2}$ dan kichik yoy aylananing chap qismida:

Javob. $x \in \left[\tfrac{3\pi}{4}+2\pi n;\ \tfrac{5\pi}{4}+2\pi n\right]$.

Eslatma. Sinus tengsizligida yoy gorizontal chiziqqa nisbatan (yuqori/past), kosinusda esa vertikal chiziqqa nisbatan (o‘ng/chap) belgilanadi. Bu farqni aylanada ko‘rib olish xatoning oldini oladi.

§3. Tangens va murakkabroq hollar

Misol. $\tan x \le 1$ tengsizlikni bir davrda yeching.
Yechish. Tangens $\left(-\tfrac{\pi}{2};\tfrac{\pi}{2}\right)$ oralig‘ida o‘suvchi. $\tan x = 1 \Rightarrow x = \tfrac{\pi}{4}$. Demak shu oraliqda $\tan x \le 1$ bo‘lishi uchun:

Quyi chegara ochiq, chunki $x=-\tfrac{\pi}{2}$ da tangens aniqlanmagan.

Javob. $x \in \left(-\tfrac{\pi}{2}+\pi n;\ \tfrac{\pi}{4}+\pi n\right]$.

Misol. $2\sin^2 x - 1 > 0$ tengsizlikni yeching.
Yechish. $2\sin^2 x - 1 = -\cos 2x$ (chunki $\cos 2x = 1 - 2\sin^2 x$). Demak $-\cos 2x > 0$, ya’ni $\cos 2x < 0$:

Ikkiga bo‘lamiz:

Javob. $x \in \left(\tfrac{\pi}{4}+\pi n;\ \tfrac{3\pi}{4}+\pi n\right)$.

Eslatma. Argumenti $2x$ bo‘lganda davr ham o‘zgaradi: yechimni $2x$ uchun topib, so‘ng ikkiga bo‘lganda davr $2\pi$ dan $\pi$ ga qisqaradi.

§ Lug‘at

trigonometrik tengsizlik — trigonometric inequality
birlik aylana — unit circle
ordinata — ordinate
abssissa — abscissa
davriylik — periodicity
yoy — arc