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^4(x)+sin^2(x)-1=0

解

2 sin^{4} (x) + sin^{2} (x) - 1 = 0

解

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{4} (x) + sin^{2} (x) - 1 = 0

置換で解く

2 sin^{4} (x) + sin^{2} (x) - 1 = 0

仮定:

sin (x) = u

2 u^{4} + u^{2} - 1 = 0

2 u^{4} + u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = i, u = - i

2 u^{4} + u^{2} - 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

2 v^{2} + v - 1 = 0

解く

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

2 v^{2} + v - 1 = 0

解くとthe二次式

2 v^{2} + v - 1 = 0

二次Equationの公式:

次の場合:

a = 2, b = 1, c = - 1

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1) = 3

1^{2} - 4 \cdot 2 (- 1)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

v_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

解を分離する

v_{1} = \frac{- 1 + 3}{2 \cdot 2}, v_{2} = \frac{- 1 - 3}{2 \cdot 2}

v = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

数を足す/引く:

- 1 + 3 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

v = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

数を引く:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

v = \frac{1}{2}, v = - 1

v = \frac{1}{2}, v = - 1

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

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

u^{2} = \frac{1}{2}

x^{2} = f (a)

の場合, 解は

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

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

解く

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

x^{2} = f (a)

の場合, 解は

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

u = - 1, u = - - 1

簡素化

- 1 : i

- 1

虚数の規則を適用する:

- 1 = i

= i

簡素化

- - 1 : - i

- - 1

虚数の規則を適用する:

- 1 = i

= - i

u = i, u = - i

解答は

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

代用を戻す

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

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{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = i :

解なし

sin (x) = i

解なし

sin (x) = - i :

解なし

sin (x) = - i

解なし

すべての解を組み合わせる

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

グラフ

人気の例

tan(x/9)+sqrt(3)=0

tan (\frac{x}{9}) + 3 = 0

cos (θ) = \frac{6}{9}

sec (b) = 2

2sin^2(x)-4sin(x)=cos^2(x)-2

2 sin^{2} (x) - 4 sin (x) = cos^{2} (x) - 2

cos(θ/2)=-(sqrt(2))/2

cos (\frac{θ}{2}) = - \frac{2}{2}