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人気のある三角関数 >

3sin^4(x)+cos^4(x)=1

解

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

解

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

1

を引く

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

指数の規則を適用する:

a^{b} = a^{2} a^{b - 2}

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

簡素化

- 1 + cos^{4} (x) + 3 (1 - cos^{2} (x)) (1 - cos^{2} (x)) : 4 cos^{4} (x) - 6 cos^{2} (x) + 2

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

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

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

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

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

数を足す:

1 + 1 = 2

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

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

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

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

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

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

簡素化

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

2 \cdot 1 \cdot cos^{2} (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos^{2} (x)

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

(cos^{2} (x))^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (x)

数を乗じる:

2 \cdot 2 = 4

= cos^{4} (x)

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

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

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

拡張

3 (1 - 2 cos^{2} (x) + cos^{4} (x)) : 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

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

括弧を分配する

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

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

3 \cdot 1 - 3 \cdot 2 cos^{2} (x) + 3 cos^{4} (x) : 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

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

数を乗じる:

3 \cdot 1 = 3

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

数を乗じる:

3 \cdot 2 = 6

= 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

= 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

= - 1 + cos^{4} (x) + 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

簡素化

- 1 + cos^{4} (x) + 3 - 6 cos^{2} (x) + 3 cos^{4} (x) : 4 cos^{4} (x) - 6 cos^{2} (x) + 2

- 1 + cos^{4} (x) + 3 - 6 cos^{2} (x) + 3 cos^{4} (x)

条件のようなグループ

= cos^{4} (x) - 6 cos^{2} (x) + 3 cos^{4} (x) - 1 + 3

類似した元を足す:

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

= 4 cos^{4} (x) - 6 cos^{2} (x) - 1 + 3

数を足す/引く:

- 1 + 3 = 2

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

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

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

2 + 4 cos^{4} (x) - 6 cos^{2} (x) = 0

置換で解く

2 + 4 cos^{4} (x) - 6 cos^{2} (x) = 0

仮定:

cos (x) = u

2 + 4 u^{4} - 6 u^{2} = 0

2 + 4 u^{4} - 6 u^{2} = 0 : u = 1, u = - 1, u = \frac{1}{2}, u = - \frac{1}{2}

2 + 4 u^{4} - 6 u^{2} = 0

標準的な形式で書く

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

4 u^{4} - 6 u^{2} + 2 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

4 v^{2} - 6 v + 2 = 0

解く

4 v^{2} - 6 v + 2 = 0 : v = 1, v = \frac{1}{2}

4 v^{2} - 6 v + 2 = 0

解くとthe二次式

4 v^{2} - 6 v + 2 = 0

二次Equationの公式:

次の場合:

a = 4, b = - 6, c = 2

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 \cdot 2 = 2

(- 6)^{2} - 4 \cdot 4 \cdot 2

指数の規則を適用する:

n

が偶数であれば

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

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 4 \cdot 2

数を乗じる:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

数を引く:

36 - 32 = 4

= 4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

2^{2} = 2

= 2

v_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 4}

解を分離する

v_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 4}, v_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 4}

v = \frac{- ( - 6 ) + 2}{2 \cdot 4} : 1

\frac{- ( - 6 ) + 2}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{6 + 2}{2 \cdot 4}

数を足す:

6 + 2 = 8

= \frac{8}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= 1

v = \frac{- ( - 6 ) - 2}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 6 ) - 2}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{6 - 2}{2 \cdot 4}

数を引く:

6 - 2 = 4

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

数を乗じる:

2 \cdot 4 = 8

= \frac{4}{8}

共通因数を約分する:

4

= \frac{1}{2}

二次equationの解:

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

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

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

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

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

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}

解答は

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

代用を戻す

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

以下の一般解

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

10進法形式で解を証明する

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

グラフ

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