Trigonometrik tenglamalar

Bu nima va nega kerak

Mayatnik tebranishi, o'zgaruvchan tok va tovush to'lqinlari kabi davriy jarayonlar trigonometrik funksiyalar bilan ifodalanadi.

Mayatnik og'ishi $x(t) = 10\cos(2t)$ formulasi bilan berilgan bo'lsa, uning qachon $5\text{ sm}$ masofada bo'lishini topish uchun $\cos(2t) = 0{,}5$ tenglamani yechish kerak.

Bunday tenglamalar trigonometrik tenglamalar deyiladi. Funksiyalarning davriyligi tufayli ularning yechimlari cheksiz ko'p bo'ladi. Ushbu darsda eng sodda trigonometrik tenglamalar va ularni yechish usullarini o'rganamiz.

Asosiy g'oya

Noma'lum burchak trigonometrik funksiya belgisi ostida kelgan tenglama trigonometrik tenglama deyiladi.

1. Eng sodda trigonometrik tenglamalar ($k \in \mathbb{Z}$)

1. $\sin x = a$:

• $|a| \le 1$ da: $x = (-1)^k \arcsin a + \pi k$.
• Xususiy hollar: $\sin x = 0 \implies x = \pi k$; $\sin x = 1 \implies x = \frac{\pi}{2} + 2\pi k$; $\sin x = -1 \implies x = -\frac{\pi}{2} + 2\pi k$.
• Xossa: $\arcsin(-a) = -\arcsin a$. $|a| > 1$ da ildiz yo'q.

2. $\cos x = a$:

• $|a| \le 1$ da: $x = \pm \arccos a + 2\pi k$.
• Xususiy hollar: $\cos x = 0 \implies x = \frac{\pi}{2} + \pi k$; $\cos x = 1 \implies x = 2\pi k$; $\cos x = -1 \implies x = \pi + 2\pi k$.
• Xossa: $\arccos(-a) = \pi - \arccos a$. $|a| > 1$ da ildiz yo'q.

3. $\operatorname{tg} x = a$:

• Har qanday $a \in \mathbb{R}$ da: $x = \operatorname{arctg} a + \pi k$.
• Xossa: $\operatorname{arctg}(-a) = -\operatorname{arctg} a$.

4. $\operatorname{ctg} x = a$:

• Har qanday $a \in \mathbb{R}$ da: $x = \operatorname{arcctg} a + \pi k$.
• Xossa: $\operatorname{arcctg}(-a) = \pi - \operatorname{arcctg} a$.

2. Yechishning asosiy usullari

1. Yangi o'zgaruvchi kiritish: $t = \sin x$ yoki $t = \cos x$ orqali kvadrat tenglamaga keltiriladi ($|t| \le 1$).
2. Bir jinsli tenglamalar: $a\sin x + b\cos x = 0$ ikkala tomonini $\cos x \neq 0$ ga bo'lib, $\operatorname{tg} x = -\frac{b}{a}$ hosil qilinadi.
3. Ko'paytuvchilarga ajratish: $f(x)\cdot g(x) = 0$ shakliga keltirilib, har bir ko'paytuvchi nolga tenglanadi.
4. Yordamchi burchak kiritish: $a\sin x + b\cos x = c$ tenglama ikkala tomoni $\sqrt{a^2 + b^2}$ ga bo'linadi (yechim sharti: $|c| \le \sqrt{a^2 + b^2}$).

Yechilgan misollar

1-misol. Eng sodda tenglamalar

Tenglamalarni yeching: a) $\sin x = -\dfrac{1}{2}$; b) $\cos 2x = \dfrac{\sqrt{2}}{2}$.

• a) $\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$ bo'lgani uchun: $x = (-1)^k\left(-\frac{\pi}{6}\right) + \pi k = (-1)^{k+1}\frac{\pi}{6} + \pi k, \quad k \in \mathbb{Z}.$
• b) Kosinus formulasidan: $2x = \pm \frac{\pi}{4} + 2\pi k \implies x = \pm \frac{\pi}{8} + \pi k, \quad k \in \mathbb{Z}.$

Javob: a) $x = (-1)^{k+1}\dfrac{\pi}{6} + \pi k$; b) $x = \pm \dfrac{\pi}{8} + \pi k, \; k \in \mathbb{Z}$.

2-misol. Kvadrat tenglamaga keltirish

$2\cos^2 x - 5\cos x + 2 = 0$

tenglamani yeching.

• $\cos x = t$ deb belgilaymiz ($|t| \le 1$): $2t^2 - 5t + 2 = 0 \implies t_1 = 2, \quad t_2 = \frac{1}{2}.$
• $t_1 = 2 > 1$ bo'lgani sababli yechim yo'q.
• $t_2 = \dfrac{1}{2}$ dan: $\cos x = \frac{1}{2} \implies x = \pm \frac{\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}.$

Javob: $x = \pm \dfrac{\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$.

3-misol. Bir jinsli tenglama

$\sin^2 x - 4\sin x\cos x + 3\cos^2 x = 0$

tenglamani yeching.

• $\cos x \neq 0$ bo'lgani uchun ikkala tomonni $\cos^2 x$ ga bo'lamiz: $\operatorname{tg}^2 x - 4\operatorname{tg} x + 3 = 0.$
• Bundan $\operatorname{tg} x = 1$ yoki $\operatorname{tg} x = 3$:
  1. $\operatorname{tg} x = 1 \implies x = \dfrac{\pi}{4} + \pi k, \quad k \in \mathbb{Z}$.
  2. $\operatorname{tg} x = 3 \implies x = \operatorname{arctg} 3 + \pi n, \quad n \in \mathbb{Z}$.

Javob: $x = \dfrac{\pi}{4} + \pi k$ va $x = \operatorname{arctg} 3 + \pi n, \quad k, n \in \mathbb{Z}$.

4-misol. Berilgan kesmadagi ildizlar

$\sin 3x = 0$ tenglamaning $[0; \pi]$ kesmadagi barcha ildizlarini toping.

• Umumiy yechim: $3x = \pi k \implies x = \dfrac{\pi k}{3}, \quad k \in \mathbb{Z}$.
• $[0; \pi]$ kesmada yotish sharti: $0 \le \frac{\pi k}{3} \le \pi \implies 0 \le k \le 3.$
• $k \in \{0, 1, 2, 3\}$ qiymatlariga mos ildizlar: $x_1 = 0$, $x_2 = \dfrac{\pi}{3}$, $x_3 = \dfrac{2\pi}{3}$, $x_4 = \pi$.

Javob: $0; \; \dfrac{\pi}{3}; \; \dfrac{2\pi}{3}; \; \pi$ (jami 4 ta ildiz).

Mashq

Quyidagi topshiriqlarni mustaqil bajaring:

1. $\sin x = \dfrac{\sqrt{3}}{2}$ tenglamaning eng kichik musbat ildizini toping.
2. $\cos\left(x - \dfrac{\pi}{4}\right) = 0$ tenglamani yeching.
3. $\operatorname{tg} 3x = \sqrt{3}$ tenglamaning umumiy yechimini yozing.
4. $2\sin^2 x - 3\sin x + 1 = 0$ tenglamani yeching.
5. $\sin x + \cos x = 0$ tenglamaning umumiy yechimini toping.
6. $\cos 2x = -1$ tenglamaning $[0; 2\pi]$ kesmadagi ildizlarini toping.
7. $2\cos^2 x + \cos x - 1 = 0$ ning $[0; \pi]$ dagi eng katta ildizini aniqlang.
8. $\sin 2x = \cos x$ tenglamani yeching.
9. $\sqrt{3}\operatorname{tg} x - 3 = 0$ ning $(0; 2\pi)$ oraliqdagi ildizlari yig'indisini toping.
10. $\sin^2 x - 3\cos x + 3 = 0$ tenglamani yeching.

Javoblar va yechim yo'llari

1. $x = (-1)^k \dfrac{\pi}{3} + \pi k$. Eng kichik musbat ildiz ($k = 0$ da): $\dfrac{\pi}{3}$. Javob: $\dfrac{\pi}{3}$.

2. $x - \dfrac{\pi}{4} = \dfrac{\pi}{2} + \pi k \implies x = \dfrac{3\pi}{4} + \pi k$. Javob: $x = \dfrac{3\pi}{4} + \pi k, \; k \in \mathbb{Z}$.

3. $3x = \dfrac{\pi}{3} + \pi k \implies x = \dfrac{\pi}{9} + \dfrac{\pi k}{3}$. Javob: $x = \dfrac{\pi}{9} + \dfrac{\pi k}{3}, \; k \in \mathbb{Z}$.

4. $2t^2 - 3t + 1 = 0 \implies t_1 = 1, \; t_2 = \dfrac{1}{2}$. Javob: $x = \dfrac{\pi}{2} + 2\pi k$ va $x = (-1)^n \dfrac{\pi}{6} + \pi n, \; k, n \in \mathbb{Z}$.

5. $\cos x$ ga bo'lsak: $\operatorname{tg} x = -1 \implies x = -\dfrac{\pi}{4} + \pi k$. Javob: $x = -\dfrac{\pi}{4} + \pi k, \; k \in \mathbb{Z}$.

6. $2x = \pi + 2\pi k \implies x = \dfrac{\pi}{2} + \pi k$. $[0; 2\pi]$ da: $\dfrac{\pi}{2}$ va $\dfrac{3\pi}{2}$. Javob: $\dfrac{\pi}{2}$ va $\dfrac{3\pi}{2}$.

7. $2t^2 + t - 1 = 0 \implies t_1 = \dfrac{1}{2}, \; t_2 = -1$. $[0; \pi]$ da: $x_1 = \dfrac{\pi}{3}$, $x_2 = \pi$. Javob: $\pi$.

8. $2\sin x\cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0$. Javob: $x = \dfrac{\pi}{2} + \pi k$ va $x = (-1)^n \dfrac{\pi}{6} + \pi n, \; k, n \in \mathbb{Z}$.

9. $\operatorname{tg} x = \sqrt{3} \implies x_1 = \dfrac{\pi}{3}, \; x_2 = \dfrac{4\pi}{3}$. Yig'indi: $\dfrac{5\pi}{3}$. Javob: $\dfrac{5\pi}{3}$.

10. $1 - \cos^2 x - 3\cos x + 3 = 0 \implies \cos^2 x + 3\cos x - 4 = 0 \implies \cos x = 1$. Javob: $x = 2\pi k, \; k \in \mathbb{Z}$.

Keng tarqalgan xatolar

1. Qiymatlar sohasidan tashqaridagi songa teskari funksiya qo'llash. ✗ $\sin x = 1{,}5$ tenglamadan $x = \arcsin(1{,}5)$ deb yozish. ✓ $|\sin x| \le 1$ va $|\cos x| \le 1$ bo'lgani uchun $|a| > 1$ da tenglama yechimga ega emas.

2. Davrni yozishni unutish. ✗ $\sin x = \dfrac{1}{2}$ tenglamaning javobini faqat $x = \dfrac{\pi}{6}$ deb yakunlash. ✓ Trigonometrik funksiyalar davriydir, doimo davr qo'shiladi: $x = (-1)^k \dfrac{\pi}{6} + \pi k, \; k \in \mathbb{Z}$.

3. Manfiy sonning arkkosinusida adashish. ✗ $\arccos\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{3}$ deb manfiy qiymat olish. ✓ Kosinus uchun qoida: $\arccos(-a) = \pi - \arccos a$. Demak, $\arccos\left(-\dfrac{1}{2}\right) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.

4. O'zgaruvchili ifodaga bo'lib, ildiz yo'qotish. ✗ $\sin 2x = \cos x$ tenglamani $\cos x$ ga qisqartirib yuborish. ✓ $\cos x = 0$ bo'lishi mumkin! Hadlarni bir tomonga o'tkazib, $\cos x$ ni qavsdan tashqariga chiqarish kerak.

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Keyingi qadam

Trigonometrik tenglamalarni yechishni o'rgandik. Navbatdagi muhim mavzu — trigonometrik funksiyalar ma'lum oraliqlarda yotganda argumentni topishga bag'ishlangan Trigonometrik tengsizliklar.