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人気のある三角関数 >

2cot^4(x)-cot^2(x)-15=0

解

2 cot^{4} (x) - cot^{2} (x) - 15 = 0

解

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

+1

度

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

解答ステップ

2 cot^{4} (x) - cot^{2} (x) - 15 = 0

置換で解く

2 cot^{4} (x) - cot^{2} (x) - 15 = 0

仮定:

cot (x) = u

2 u^{4} - u^{2} - 15 = 0

2 u^{4} - u^{2} - 15 = 0 : u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

2 u^{4} - u^{2} - 15 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

2 v^{2} - v - 15 = 0

解く

2 v^{2} - v - 15 = 0 : v = 3, v = - \frac{5}{2}

2 v^{2} - v - 15 = 0

解くとthe二次式

2 v^{2} - v - 15 = 0

二次Equationの公式:

次の場合:

a = 2, b = - 1, c = - 15

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 15) = 11

(- 1)^{2} - 4 \cdot 2 (- 15)

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 15

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

(- 1)^{2} = 1^{2}

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 2 \cdot 15 = 120

4 \cdot 2 \cdot 15

数を乗じる:

4 \cdot 2 \cdot 15 = 120

= 120

= 1 + 120

数を足す:

1 + 120 = 121

= 121

数を因数に分解する:

121 = 1 1^{2}

= 1 1^{2}

累乗根の規則を適用する:

n a^{n} = a

1 1^{2} = 11

= 11

v_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 2}

解を分離する

v_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 2}, v_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 2}

v = \frac{- ( - 1 ) + 11}{2 \cdot 2} : 3

\frac{- ( - 1 ) + 11}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{1 + 11}{2 \cdot 2}

数を足す:

1 + 11 = 12

= \frac{12}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{12}{4}

数を割る:

\frac{12}{4} = 3

= 3

v = \frac{- ( - 1 ) - 11}{2 \cdot 2} : - \frac{5}{2}

\frac{- ( - 1 ) - 11}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{1 - 11}{2 \cdot 2}

数を引く:

1 - 11 = - 10

= \frac{- 10}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 10}{4}

分数の規則を適用する:

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

= - \frac{10}{4}

共通因数を約分する:

2

= - \frac{5}{2}

二次equationの解:

v = 3, v = - \frac{5}{2}

v = 3, v = - \frac{5}{2}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

x^{2} = f (a)

の場合, 解は

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

u = 3, u = - 3

解く

u^{2} = - \frac{5}{2} : u = i \frac{5}{2}, u = - i \frac{5}{2}

u^{2} = - \frac{5}{2}

x^{2} = f (a)

の場合, 解は

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

u = - \frac{5}{2}, u = - - \frac{5}{2}

簡素化

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

累乗根の規則を適用する:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{5}{2}

簡素化

- - \frac{5}{2} : - i \frac{5}{2}

- - \frac{5}{2}

簡素化

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

累乗根の規則を適用する:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

虚数の規則を適用する:

- 1 = i

= i \frac{5}{2}

= - i \frac{5}{2}

u = i \frac{5}{2}, u = - i \frac{5}{2}

解答は

u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

代用を戻す

u = cot (x)

cot (x) = 3, cot (x) = - 3, cot (x) = i \frac{5}{2}, cot (x) = - i \frac{5}{2}

cot (x) = 3, cot (x) = - 3, cot (x) = i \frac{5}{2}, cot (x) = - i \frac{5}{2}

cot (x) = 3 : x = \frac{π}{6} + πn

cot (x) = 3

以下の一般解

cot (x) = 3

cot (x)

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

cot (x) = - 3 : x = \frac{5 π}{6} + πn

cot (x) = - 3

以下の一般解

cot (x) = - 3

cot (x)

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

cot (x) = i \frac{5}{2} :

解なし

cot (x) = i \frac{5}{2}

解なし

cot (x) = - i \frac{5}{2} :

解なし

cot (x) = - i \frac{5}{2}

解なし

すべての解を組み合わせる

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

グラフ

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