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 >

2sin^2(x)+sin(x)=0,0<= x<= 2pi

解

2 sin^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π

解

x = 0, x = π, x = 2 π, x = \frac{7 π}{6}, x = \frac{11 π}{6}

+1

度

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

解答ステップ

2 sin^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π

置換で解く

2 sin^{2} (x) + sin (x) = 0

仮定:

sin (x) = u

2 u^{2} + u = 0

2 u^{2} + u = 0 : u = 0, u = - \frac{1}{2}

2 u^{2} + u = 0

解くとthe二次式

2 u^{2} + u = 0

二次Equationの公式:

次の場合:

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 1 - 0

数を引く:

1 - 0 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 1 + 1}{2 \cdot 2}, u_{2} = \frac{- 1 - 1}{2 \cdot 2}

u = \frac{- 1 + 1}{2 \cdot 2} : 0

\frac{- 1 + 1}{2 \cdot 2}

数を足す/引く:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{4}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 1 - 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 1 - 1}{2 \cdot 2}

数を引く:

- 1 - 1 = - 2

= \frac{- 2}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{4}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

二次equationの解:

u = 0, u = - \frac{1}{2}

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = - \frac{1}{2}

sin (x) = 0, sin (x) = - \frac{1}{2}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

以下の一般解

sin (x) = 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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

以下の一般解

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

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 = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

すべての解を組み合わせる

x = 0, x = π, x = 2 π, x = \frac{7 π}{6}, x = \frac{11 π}{6}

グラフ

人気の例

6cos^2(x)-5cos(x)=4

6 cos^{2} (x) - 5 cos (x) = 4

cos (2 0^{\circ}) = sin (x)

sec (2 x) = 1

2cos^2(x)+sin^2(x)=0

2 cos^{2} (x) + sin^{2} (x) = 0

3cos(2x)-5sin(x)+1=0

3 cos (2 x) - 5 sin (x) + 1 = 0