23. Муодилаи sin x = a - Алгебра — синфи 10
Решаҳои sin x = a \sin x = a sin x = a
∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1 ∣ sin x ∣ ≤ 1 |\sin x| \le 1 ∣ sin x ∣ ≤ 1 x . x. x .
навбатӣ 24. Муодилаи cos x = a 1 соат
∣ a ∣ ≤ 1 |a| \le 1 ∣ a ∣ ≤ 1 [ − π 2 ; π 2 ] \left[-\dfrac{\pi}{2};\ \dfrac{\pi}{2}\right] [ − 2 π ; 2 π ] a a a x 1 = arcsin a . x_1 = \arcsin a. x 1 = arcsin a . [ π 2 ; 3 π 2 ] \left[\dfrac{\pi}{2};\ \dfrac{3\pi}{2}\right] [ 2 π ; 2 3 π ] 1 1 1 − 1 -1 − 1 a a a x 2 = π − arcsin a : x_2 = \pi - \arcsin a: x 2 = π − arcsin a : sin ( π − x 1 ) = sin x 1 = a \sin(\pi - x_1) = \sin x_1 = a sin ( π − x 1 ) = sin x 1 = a − π 2 ≤ x 1 ≤ π 2 -\dfrac{\pi}{2} \le x_1 \le \dfrac{\pi}{2} − 2 π ≤ x 1 ≤ 2 π π 2 ≤ π − x 1 ≤ 3 π 2 . \dfrac{\pi}{2} \le \pi - x_1 \le \dfrac{3\pi}{2}. 2 π ≤ π − x 1 ≤ 2 Пас, дар порчаи дарозиаш 2 π 2\pi 2 π − π 2 -\dfrac{\pi}{2} − 2 π 3 π 2 \dfrac{3\pi}{2} 2 3 π 2 π 2\pi 2 π x = arcsin a + 2 π n , x = π − arcsin a + 2 π n , n ∈ Z . x = \arcsin a + 2\pi n, \qquad x = \pi - \arcsin a + 2\pi n, \qquad n \in \mathbb{Z}. x = arcsin a + 2 πn , x = π − arcsin a + 2 πn , n ∈ Z .
Дар давраи воҳидӣ: sin x \sin x sin x P x . P_x. P x . ∣ a ∣ < 1 |a| < 1 ∣ a ∣ < 1 ∣ a ∣ = 1 |a| = 1 ∣ a ∣ = 1
Мисоли 1. sin 2 x 3 = 1 2 \sin\dfrac{2x}{3} = \dfrac{1}{2} sin 3 2 x = 2 1
Ҳал. 2 x 3 = ( − 1 ) n arcsin 1 2 + π n = ( − 1 ) n π 6 + π n , \dfrac{2x}{3} = (-1)^n\arcsin\dfrac{1}{2} + \pi n = (-1)^n\dfrac{\pi}{6} + \pi n, 3 2 x = ( − 1 ) n arcsin 2 1 + πn = ( − 1 ) n 6 π + πn , x = ( − 1 ) n π 4 + 3 π n 2 , n ∈ Z . x = (-1)^n\frac{\pi}{4} + \frac{3\pi n}{2}, \qquad n \in \mathbb{Z}. x = ( − 1 ) n 4 π + 2
Мисоли 2. sin 2 π x = 2 2 \sin\dfrac{2\pi}{x} = \dfrac{\sqrt{2}}{2} sin x 2 π = 2 2
Ҳал. 2 π x = ( − 1 ) n π 4 + π n , \dfrac{2\pi}{x} = (-1)^n\dfrac{\pi}{4} + \pi n, x 2 π = ( − 1 ) n 4 π + πn , π \pi π 2 x = ( − 1 ) n 4 + n = 4 n + ( − 1 ) n 4 . \dfrac{2}{x} = \dfrac{(-1)^n}{4} + n = \dfrac{4n + (-1)^n}{4}. x 2 = 4 ( − 1 ) n x = 8 4 n + ( − 1 ) n , n ∈ Z . x = \frac{8}{4n + (-1)^n}, \qquad n \in \mathbb{Z}. x = 4 n + ( − 1 ) n 8 , n ∈ Z . 4 n ± 1 ≠ 0 4n \pm 1 \ne 0 4 n ± 1 = 0 n ∈ Z . n \in \mathbb{Z}. n ∈ Z .
Мисоли 3 (★). sin 3 π x = − 3 2 \sin\dfrac{3\pi}{\sqrt{x}} = -\dfrac{\sqrt{3}}{2} sin x 3 π = − 2 3
Ҳал. Азбаски arcsin ( − 3 2 ) = − π 3 , \arcsin\left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{\pi}{3}, arcsin ( − 2 3 ) = − 3 π , 3 π x = ( − 1 ) n + 1 π 3 + π n . \dfrac{3\pi}{\sqrt{x}} = (-1)^{n+1}\dfrac{\pi}{3} + \pi n. x 3 π = ( π \pi π 3 x = ( − 1 ) n + 1 3 + n = 3 n + ( − 1 ) n + 1 3 , \dfrac{3}{\sqrt{x}} = \dfrac{(-1)^{n+1}}{3} + n = \dfrac{3n + (-1)^{n+1}}{3}, x 3 x = 9 3 n + ( − 1 ) n + 1 . \sqrt{x} = \dfrac{9}{3n + (-1)^{n+1}}. x = 3 n + ( − 1 ) x > 0 \sqrt{x} > 0 x > 0 3 n + ( − 1 ) n + 1 > 0. 3n + (-1)^{n+1} > 0. 3 n + ( − 1 ) n + 1 > 0. n ≤ 0 n \le 0 n ≤ 0 (n = 0 n = 0 n = 0 медиҳад − 1 -1 − 1 n ≥ 1 n \ge 1 n ≥ 1 x = 81 ( 3 n + ( − 1 ) n + 1 ) 2 , n = 1 , 2 , 3 , … x = \frac{81}{\left(3n + (-1)^{n+1}\right)^2}, \qquad n = 1,\ 2,\ 3,\ \ldots x = ( 3 n + ( − 1 ) n + 1 ) 2 n = 1 n = 1 n = 1 x = 9 4 , \sqrt{x} = \dfrac{9}{4}, x = 4 9 , 3 π 9 / 4 = 4 π 3 , \dfrac{3\pi}{9/4} = \dfrac{4\pi}{3}, 9/4 3 π = 3 4 π , sin 4 π 3 = − 3 2 . \sin\dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}. sin 3 4 π = − 2 3
Худатон санҷед: a a a n n n
sin x = a: ҳамаи ҳалҳо санҷида бин! Аз нав
Кор дар синф
Машқҳо
Дар синф
а) sin 3 x 4 = 3 2 ; \sin\dfrac{3x}{4} = \dfrac{\sqrt{3}}{2}; sin 4 3 x = 2 3 ;
б) sin 2 π x = − 1 2 ; \sin\dfrac{2\pi}{x} = -\dfrac{1}{2}; sin x 2 π = − 2 1 ;
в) sin 4 π x 2 = 1 ; \sin\dfrac{4\pi}{x^2} = 1; sin x 2 4 π = 1 ;
г) sin 3 π x = − 1 ; \sin\dfrac{3\pi}{\sqrt{x}} = -1; sin x 3 π = − 1 ;
д) sin π x = 0 ; \sin\sqrt{\dfrac{\pi}{x}} = 0; sin x π = 0 ;
е) sin ( 3 − 2 x ) = − 2 2 ; \sin(3 - 2x) = -\dfrac{\sqrt{2}}{2}; sin ( 3 − 2 x ) = − 2 2
ж) sin 2 x = π 4 ; \sin 2x = \dfrac{\pi}{4}; sin 2 x = 4 π ;
з) sin x = π 3 ; \sin x = \dfrac{\pi}{3}; sin x = 3 π ;
и) sin x = 0,01 . \sin x = \sqrt{0{,}01}. sin x = 0 , 01 .
а) sin 4 x = − 1 ; \sin 4x = -1; sin 4 x = − 1 ;
б) sin x = − 2 2 ; \sin x = -\dfrac{\sqrt{2}}{2}; sin x = − 2 2 ;
в) sin 2 x = 1 2 ; \sin 2x = \dfrac{1}{2}; sin 2 x = 2 1 ;
г) sin 4 x = 1 ; \sin 4x = 1; sin 4 x = 1 ;
д) 2 sin x = 2 ; 2\sin x = \sqrt{2}; 2 sin x = 2 ;
е) 2 sin 2 x = − 1 ; 2\sin 2x = -1; 2 sin 2 x = − 1 ;
ж) sin x ( 2 sin x − 2 ) = 0. \sin x\left(2\sin x - \sqrt{2}\right) = 0. sin x ( 2 sin x − 2 ) = 0.
Дар хона 3. Системаи муодилаҳоро ҳал кунед:
а) { 3 x + y + 4 = 0 , x 2 − y 2 = 2 ; \begin{cases} 3x + y + 4 = 0, \\ x^2 - y^2 = 2; \end{cases} { 3 x + y + 4 = 0 , x 2 − y 2 = 2 ;
б) { y + 3 x = 2 , x 2 − x y = 3,36. \begin{cases} y + 3x = 2, \\ x^2 - xy = 3{,}36. \end{cases} { y + 3 x = 2 , x 2 − x y = 3 , 36.
4. Агар сурати касри одӣ ба квадрат бардошта шавад ва махраҷаш як воҳид кам карда шавад, касри ба адади 2 2 2 1 1 1 1 1 1 1 4 \dfrac{1}{4} 4 1
5. Ифодаи sin α 1 + cos α + sin α 1 − cos α \dfrac{\sin\alpha}{1 + \cos\alpha} + \dfrac{\sin\alpha}{1 - \cos\alpha} 1 + cos α sin α + 1 − cos α sin α sin α = − 1 8 \sin\alpha = -\dfrac{1}{8} sin α = − 8 1
Вазифаи хонагӣ
Машқҳои 3–5 ва қисми иҷронашудаи машқҳои «Дар синф».
Вазифаи хонагии онлайн: «Алгебра 10 · Вазифаи хонагӣ: 23. Муодилаи sin x = a».
Хулоса
∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1 ∣ a ∣ ≤ 1 |a| \le 1 ∣ a ∣ ≤ 1 x = ( − 1 ) n arcsin a + π n , x = (-1)^n\arcsin a + \pi n, x = ( − 1 ) n arcsin a + πn , n ∈ Z n \in \mathbb{Z} n ∈ Z arcsin a + 2 π n \arcsin a + 2\pi n arcsin a + 2 πn π − arcsin a + 2 π n \pi - \arcsin a + 2\pi n π − arcsin a + 2 πn Махсус: sin x = 1 ⇒ x = π 2 + 2 π n ; \sin x = 1 \Rightarrow x = \dfrac{\pi}{2} + 2\pi n; sin x = 1 ⇒ x = 2 π + 2 πn ; sin x = − 1 ⇒ x = − π 2 + 2 π n ; \sin x = -1 \Rightarrow x = -\dfrac{\pi}{2} + 2\pi n;
Дар sin k x = a \sin kx = a sin k x = a sin c x = a \sin\dfrac{c}{x} = a sin x c = a x x x
Дар ин дарс
Ман медонам , ки sin x = a \sin x = a sin x = a ∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1 ∣ a ∣ ≤ 1 |a| \le 1 ∣ a ∣ ≤ 1 Ман метавонам ҳалли умумиро нависам: x = ( − 1 ) n arcsin a + π n , x = (-1)^n\arcsin a + \pi n, x = ( − 1 ) n arcsin a + πn , n ∈ Z . n \in \mathbb{Z}. n ∈ Z . Ман метавонам ҳолатҳои a = 1 , a = 1, a = 1 , − 1 , -1, − 1 , 0 0 0 sin k x = a \sin kx = a sin k x = a sin c x = a \sin\dfrac{c}{x} = a sin Такрор (2 дақиқа)
arcsin 1 2 \arcsin\dfrac{1}{2} arcsin 2 1 arcsin ( − 3 2 ) \arcsin\left(-\dfrac{\sqrt{3}}{2}\right) arcsin ( − 2 3 ) Дар порчаи [ 0 ; 2 π ] [0;\ 2\pi] [ 0 ; 2 π ] 1 2 \dfrac{1}{2} 2 1
Давраи синус чӣ қадар аст?
Ҷавобҳо
π 6 ; \dfrac{\pi}{6}; 6 π ; − π 3 . -\dfrac{\pi}{3}. − 3 π . π 6 \dfrac{\pi}{6} Вазъият барои андешаронӣ Муодилаи sin x = 1 2 \sin x = \dfrac{1}{2} sin x = 2 1 π 6 , \dfrac{\pi}{6}, 6 π , 5 π 6 , \dfrac{5\pi}{6}, 6 5 π , 13 π 6 , \dfrac{13\pi}{6}, 6 13 π , … \ldots …
Ҳалли умумӣ Ду силсиларо бо як формула менависанд:
x = ( − 1 ) n arcsin a + π n , n ∈ Z , ∣ a ∣ ≤ 1. x = (-1)^n\arcsin a + \pi n, \qquad n \in \mathbb{Z},\ |a| \le 1. x = ( − 1 ) n arcsin a + πn , n ∈
Расми 1. sin x = a ҳангоми a = 1/2: боло — ду нуқтаи давра (x₁ ва x₂ дар як давр); поён — ҳамаи буришҳои хати y = a бо синусоида (нуқтаҳои сабз). Ҳолатҳои махсус
sin x = 1 \sin x = 1 sin x = 1 x = π 2 + 2 π n ; x = \dfrac{\pi}{2} + 2\pi n; x = 2 π + 2 πn ; Калимаҳои калидӣ ҳалли умумӣ муодилаи одитарин силсилаи решаҳо арксинус
Ман дар ин дарс чӣ омӯхтам? Ман метавонам:
✓ / ✗ бигӯям, чаро sin x = a \sin x = a sin x = a
✓ / ✗ sin x = a \sin x = a sin x = a ( − 1 ) n arcsin a + π n (-1)^n\arcsin a + \pi n ( − 1
Муодилаи sin x = 1,5 \sin x = 1{,}5 sin x = 1 , 5
sin x = 2 2 \sin x = \dfrac{\sqrt{2}}{2} sin x = 2 2 (n ∈ Z n \in \mathbb{Z}
Як ҷавобро интихоб кунед A B C D
sin 2 x = 1 \sin 2x = 1 sin 2 x = 1 (n ∈ Z n \in \mathbb{Z} n ∈ Z
Як ҷавобро интихоб кунед A x = π 2 + π n x = \dfrac{\pi}{2} + \pi n x = 2 π
B C D
3 π
.
3 πn
,
n
∈
Z .
+
n =
4 4 n + ( − 1 ) n .
−
1
) n + 1
3 π
+
πn .
=
3 ( − 1 ) n + 1 +
n =
3 3 n + ( − 1 ) n + 1 ,
n + 1
9
.
81
,
n
=
1 , 2 , 3 , …
.
;
sin x = − 1 ⇒ x = − 2 π + 2 πn ;
sin x = 0 ⇒ x = π n . \sin x = 0 \Rightarrow x = \pi n. sin x = 0 ⇒ x = πn . x > 0 x > 0 x > 0
x c
=
a
6 π
5 π 6 . \dfrac{5\pi}{6}. 6 5 π .
Z , ∣ a ∣ ≤
1.
Ҳангоми n n n n = 2 k n = 2k n = 2 k arcsin a + 2 π k ; \arcsin a + 2\pi k; arcsin a + 2 πk ; n n n n = 2 k + 1 n = 2k + 1 n = 2 k + 1 − arcsin a + π + 2 π k -\arcsin a + \pi + 2\pi k − arcsin a + π + 2 πk
sin x = − 1 \sin x = -1 sin x = − 1
x = − π 2 + 2 π n ; x = -\dfrac{\pi}{2} + 2\pi n; x = − 2 π + 2 πn ; sin x = 0 \sin x = 0 sin x = 0 x = π n x = \pi n x = πn n ∈ Z n \in \mathbb{Z} n ∈ Z Ҳангоми a = ± 1 a = \pm 1 a = ± 1 x 1 = x 2 x_1 = x_2 x 1 = x 2 a = 0 a = 0 a = 0 2 π n 2\pi n 2 πn π + 2 π n \pi + 2\pi n π + 2 πn π n \pi n πn
) n
arcsin
a
+
πn
✓ / ✗ ҳолатҳои a = 0 , a = 0, a = 0 , ± 1 \pm 1 ± 1 ∣ a ∣ > 1 |a| > 1 ∣ a ∣ > 1 n ∈ Z
x = ( − 1 ) n π 4 + 2 π n x = (-1)^n\dfrac{\pi}{4} + 2\pi n x = ( − 1 ) n 4 π + 2 πn
x = ± π 4 + 2 π n x = \pm\dfrac{\pi}{4} + 2\pi n x = ± 4 π + 2 πn
x = π 4 + 2 π n x = \dfrac{\pi}{4} + 2\pi n x = 4 π + 2 πn
x = ( − 1 ) n π 4 + π n x = (-1)^n\dfrac{\pi}{4} + \pi n x = ( − 1 ) n 4 π + πn
+
πn
x = π 4 + 2 π n x = \dfrac{\pi}{4} + 2\pi n x = 4 π + 2 πn
x = π 2 + 2 π n x = \dfrac{\pi}{2} + 2\pi n x = 2 π + 2 πn
x = π 4 + π n x = \dfrac{\pi}{4} + \pi n x = 4 π + πn
