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人気のある三角関数 >

3sin^2(x)-1=cos^4(x)

解

3 sin^{2} (x) - 1 = cos^{4} (x)

解

x = 0.72368 \dots + 2 πn, x = 2 π - 0.72368 \dots + 2 πn, x = 2.41790 \dots + 2 πn, x = - 2.41790 \dots + 2 πn

+1

度

x = 41.46431 \dots^{\circ} + 36 0^{\circ} n, x = 318.53568 \dots^{\circ} + 36 0^{\circ} n, x = 138.53568 \dots^{\circ} + 36 0^{\circ} n, x = - 138.53568 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

3 sin^{2} (x) - 1 = cos^{4} (x)

両辺から

cos^{4} (x)

を引く

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

三角関数の公式を使用して書き換える

- 1 - cos^{4} (x) + 3 sin^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 1 - cos^{4} (x) + 3 (1 - cos^{2} (x))

簡素化

- 1 - cos^{4} (x) + 3 (1 - cos^{2} (x)) : - cos^{4} (x) - 3 cos^{2} (x) + 2

- 1 - cos^{4} (x) + 3 (1 - cos^{2} (x))

拡張

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

3 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

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

数を乗じる:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (x)

= - 1 - cos^{4} (x) + 3 - 3 cos^{2} (x)

簡素化

- 1 - cos^{4} (x) + 3 - 3 cos^{2} (x) : - cos^{4} (x) - 3 cos^{2} (x) + 2

- 1 - cos^{4} (x) + 3 - 3 cos^{2} (x)

条件のようなグループ

= - cos^{4} (x) - 3 cos^{2} (x) - 1 + 3

数を足す/引く:

- 1 + 3 = 2

= - cos^{4} (x) - 3 cos^{2} (x) + 2

= - cos^{4} (x) - 3 cos^{2} (x) + 2

= - cos^{4} (x) - 3 cos^{2} (x) + 2

2 - cos^{4} (x) - 3 cos^{2} (x) = 0

置換で解く

2 - cos^{4} (x) - 3 cos^{2} (x) = 0

仮定:

cos (x) = u

2 - u^{4} - 3 u^{2} = 0

2 - u^{4} - 3 u^{2} = 0 : u = i \frac{3 + 17}{2}, u = - i \frac{3 + 17}{2}, u = \frac{17 - 3}{2}, u = - \frac{17 - 3}{2}

2 - u^{4} - 3 u^{2} = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} - 3 u^{2} + 2 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

- v^{2} - 3 v + 2 = 0

解く

- v^{2} - 3 v + 2 = 0 : v = - \frac{3 + 17}{2}, v = \frac{17 - 3}{2}

- v^{2} - 3 v + 2 = 0

解くとthe二次式

- v^{2} - 3 v + 2 = 0

二次Equationの公式:

次の場合:

a = - 1, b = - 3, c = 2

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

(- 3)^{2} - 4 (- 1) \cdot 2 = 17

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 2

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

数を足す:

9 + 8 = 17

= 17

v_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 ( - 1 )}

解を分離する

v_{1} = \frac{- ( - 3 ) + 17}{2 ( - 1 )}, v_{2} = \frac{- ( - 3 ) - 17}{2 ( - 1 )}

v = \frac{- ( - 3 ) + 17}{2 ( - 1 )} : - \frac{3 + 17}{2}

\frac{- ( - 3 ) + 17}{2 ( - 1 )}

括弧を削除する:

(- a) = - a, - (- a) = a

= \frac{3 + 17}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{3 + 17}{- 2}

分数の規則を適用する:

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

= - \frac{3 + 17}{2}

v = \frac{- ( - 3 ) - 17}{2 ( - 1 )} : \frac{17 - 3}{2}

\frac{- ( - 3 ) - 17}{2 ( - 1 )}

括弧を削除する:

(- a) = - a, - (- a) = a

= \frac{3 - 17}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{3 - 17}{- 2}

分数の規則を適用する:

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

3 - 17 = - (17 - 3)

= \frac{17 - 3}{2}

二次equationの解:

v = - \frac{3 + 17}{2}, v = \frac{17 - 3}{2}

v = - \frac{3 + 17}{2}, v = \frac{17 - 3}{2}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = - \frac{3 + 17}{2} : u = i \frac{3 + 17}{2}, u = - i \frac{3 + 17}{2}

u^{2} = - \frac{3 + 17}{2}

x^{2} = f (a)

の場合, 解は

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

u = - \frac{3 + 17}{2}, u = - - \frac{3 + 17}{2}

簡素化

- \frac{3 + 17}{2} : i \frac{3 + 17}{2}

- \frac{3 + 17}{2}

累乗根の規則を適用する:

- a = - 1 a

- \frac{3 + 17}{2} = - 1 \frac{3 + 17}{2}

= - 1 \frac{3 + 17}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{3 + 17}{2}

簡素化

- - \frac{3 + 17}{2} : - i \frac{3 + 17}{2}

- - \frac{3 + 17}{2}

簡素化

- \frac{3 + 17}{2} : i \frac{3 + 17}{2}

- \frac{3 + 17}{2}

累乗根の規則を適用する:

- a = - 1 a

- \frac{3 + 17}{2} = - 1 \frac{3 + 17}{2}

= - 1 \frac{3 + 17}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{3 + 17}{2}

= - i \frac{3 + 17}{2}

u = i \frac{3 + 17}{2}, u = - i \frac{3 + 17}{2}

解く

u^{2} = \frac{17 - 3}{2} : u = \frac{17 - 3}{2}, u = - \frac{17 - 3}{2}

u^{2} = \frac{17 - 3}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{17 - 3}{2}, u = - \frac{17 - 3}{2}

解答は

u = i \frac{3 + 17}{2}, u = - i \frac{3 + 17}{2}, u = \frac{17 - 3}{2}, u = - \frac{17 - 3}{2}

代用を戻す

u = cos (x)

cos (x) = i \frac{3 + 17}{2}, cos (x) = - i \frac{3 + 17}{2}, cos (x) = \frac{17 - 3}{2}, cos (x) = - \frac{17 - 3}{2}

cos (x) = i \frac{3 + 17}{2}, cos (x) = - i \frac{3 + 17}{2}, cos (x) = \frac{17 - 3}{2}, cos (x) = - \frac{17 - 3}{2}

cos (x) = i \frac{3 + 17}{2} :

解なし

cos (x) = i \frac{3 + 17}{2}

解なし

cos (x) = - i \frac{3 + 17}{2} :

解なし

cos (x) = - i \frac{3 + 17}{2}

解なし

cos (x) = \frac{17 - 3}{2} : x = arccos \frac{17 - 3}{2} + 2 πn, x = 2 π - arccos \frac{17 - 3}{2} + 2 πn

cos (x) = \frac{17 - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{17 - 3}{2}

以下の一般解

cos (x) = \frac{17 - 3}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos \frac{17 - 3}{2} + 2 πn, x = 2 π - arccos \frac{17 - 3}{2} + 2 πn

x = arccos \frac{17 - 3}{2} + 2 πn, x = 2 π - arccos \frac{17 - 3}{2} + 2 πn

cos (x) = - \frac{17 - 3}{2} : x = arccos - \frac{17 - 3}{2} + 2 πn, x = - arccos - \frac{17 - 3}{2} + 2 πn

cos (x) = - \frac{17 - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{17 - 3}{2}

以下の一般解

cos (x) = - \frac{17 - 3}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos - \frac{17 - 3}{2} + 2 πn, x = - arccos - \frac{17 - 3}{2} + 2 πn

x = arccos - \frac{17 - 3}{2} + 2 πn, x = - arccos - \frac{17 - 3}{2} + 2 πn

すべての解を組み合わせる

x = arccos \frac{17 - 3}{2} + 2 πn, x = 2 π - arccos \frac{17 - 3}{2} + 2 πn, x = arccos - \frac{17 - 3}{2} + 2 πn, x = - arccos - \frac{17 - 3}{2} + 2 πn

10進法形式で解を証明する

x = 0.72368 \dots + 2 πn, x = 2 π - 0.72368 \dots + 2 πn, x = 2.41790 \dots + 2 πn, x = - 2.41790 \dots + 2 πn

グラフ

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