ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

sin(x^2)=0

解

sin (x^{2}) = 0

解

x = 2 πn, x = - 2 πn, x = π + 2 πn, x = - π + 2 πn

+1

度

x = 0^{\circ} + 143.61922 \dots^{\circ} n, x = 0^{\circ} - 143.61922 \dots^{\circ} n, x = 0^{\circ} + 175.89690 \dots^{\circ} n, x = 0^{\circ} - 175.89690 \dots^{\circ} n

解答ステップ

sin (x^{2}) = 0

以下の一般解

sin (x^{2}) = 0

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x^{2} = 0 + 2 πn, x^{2} = π + 2 πn

x^{2} = 0 + 2 πn, x^{2} = π + 2 πn

解く

x^{2} = 0 + 2 πn : x = 2 πn, x = - 2 πn

x^{2} = 0 + 2 πn

0 + 2 πn = 2 πn

x^{2} = 2 πn

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

x = 2 πn, x = - 2 πn

簡素化

2 πn : 2 πn

2 πn

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= 2 πn

簡素化

- 2 πn : - 2 πn

- 2 πn

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= - 2 πn

x = 2 πn, x = - 2 πn

解く

x^{2} = π + 2 πn : x = π + 2 πn, x = - π + 2 πn

x^{2} = π + 2 πn

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

x = π + 2 πn, x = - π + 2 πn

x = 2 πn, x = - 2 πn, x = π + 2 πn, x = - π + 2 πn

グラフ

人気の例

cos (x) = - \frac{1}{4}

sin (\frac{x}{3}) = \frac{1}{2}

2cos^2(t)+cos(t)=1

2 cos^{2} (t) + cos (t) = 1

9tan^3(x)=3tan(x)

9 tan^{3} (x) = 3 tan (x)

cos(3x)=0.5,0<x<pi

cos (3 x) = 0.5, 0 < x < π