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人気のある三角関数 >

5cos(x)cot(x)=1

解

5 cos (x) cot (x) = 1

解

x = 1.13135 \dots + 2 πn, x = π - 1.13135 \dots + 2 πn

+1

度

x = 64.82160 \dots^{\circ} + 36 0^{\circ} n, x = 115.17839 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

5 cos (x) cot (x) = 1

両辺から

1

を引く

5 cos (x) cot (x) - 1 = 0

サイン, コサインで表わす

5 cos (x) \frac{cos ( x )}{sin ( x )} - 1 = 0

簡素化

5 cos (x) \frac{cos ( x )}{sin ( x )} - 1 : \frac{5 cos ^{2} ( x ) - sin ( x )}{sin ( x )}

5 cos (x) \frac{cos ( x )}{sin ( x )} - 1

5 cos (x) \frac{cos ( x )}{sin ( x )} = \frac{5 cos ^{2} ( x )}{sin ( x )}

5 cos (x) \frac{cos ( x )}{sin ( x )}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( x ) \cdot 5 cos ( x )}{sin ( x )}

cos (x) \cdot 5 cos (x) = 5 cos^{2} (x)

cos (x) \cdot 5 cos (x)

指数の規則を適用する:

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

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

= 5 cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= 5 cos^{2} (x)

= \frac{5 cos ^{2} ( x )}{sin ( x )}

= \frac{5 cos ^{2} ( x )}{sin ( x )} - 1

元を分数に変換する:

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

= \frac{5 cos ^{2} ( x )}{sin ( x )} - \frac{1 \cdot sin ( x )}{sin ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 cos ^{2} ( x ) - 1 \cdot sin ( x )}{sin ( x )}

乗算:

1 \cdot sin (x) = sin (x)

= \frac{5 cos ^{2} ( x ) - sin ( x )}{sin ( x )}

\frac{5 cos ^{2} ( x ) - sin ( x )}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 cos^{2} (x) - sin (x) = 0

両辺に

sin (x)

を足す

5 cos^{2} (x) = sin (x)

両辺を2乗する

(5 cos^{2} (x))^{2} = sin^{2} (x)

両辺から

sin^{2} (x)

を引く

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

因数

25 cos^{4} (x) - sin^{2} (x) : (5 cos^{2} (x) + sin (x)) (5 cos^{2} (x) - sin (x))

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

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

を書き換え

(5 cos^{2} (x))^{2} - sin^{2} (x)

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

25

を書き換え

5^{2}

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

指数の規則を適用する:

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

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

= 5^{2} (cos^{2} (x))^{2} - sin^{2} (x)

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

5^{2} (cos^{2} (x))^{2} = (5 cos^{2} (x))^{2}

= (5 cos^{2} (x))^{2} - sin^{2} (x)

= (5 cos^{2} (x))^{2} - sin^{2} (x)

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

(5 cos^{2} (x))^{2} - sin^{2} (x) = (5 cos^{2} (x) + sin (x)) (5 cos^{2} (x) - sin (x))

= (5 cos^{2} (x) + sin (x)) (5 cos^{2} (x) - sin (x))

(5 cos^{2} (x) + sin (x)) (5 cos^{2} (x) - sin (x)) = 0

各部分を別個に解く

5 cos^{2} (x) + sin (x) = 0 or 5 cos^{2} (x) - sin (x) = 0

5 cos^{2} (x) + sin (x) = 0 : x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

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

三角関数の公式を使用して書き換える

sin (x) + 5 cos^{2} (x)

ピタゴラスの公式を使用する:

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

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

= sin (x) + 5 (1 - sin^{2} (x))

sin (x) + (1 - sin^{2} (x)) \cdot 5 = 0

置換で解く

sin (x) + (1 - sin^{2} (x)) \cdot 5 = 0

仮定:

sin (x) = u

u + (1 - u^{2}) \cdot 5 = 0

u + (1 - u^{2}) \cdot 5 = 0 : u = - \frac{- 1 + 101}{10}, u = \frac{1 + 101}{10}

u + (1 - u^{2}) \cdot 5 = 0

拡張

u + (1 - u^{2}) \cdot 5 : u + 5 - 5 u^{2}

u + (1 - u^{2}) \cdot 5

= u + 5 (1 - u^{2})

拡張

5 (1 - u^{2}) : 5 - 5 u^{2}

5 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = 5, b = 1, c = u^{2}

= 5 \cdot 1 - 5 u^{2}

数を乗じる:

5 \cdot 1 = 5

= 5 - 5 u^{2}

= u + 5 - 5 u^{2}

u + 5 - 5 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- 5 u^{2} + u + 5 = 0

解くとthe二次式

- 5 u^{2} + u + 5 = 0

二次Equationの公式:

次の場合:

a = - 5, b = 1, c = 5

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

1^{2} - 4 (- 5) \cdot 5 = 101

1^{2} - 4 (- 5) \cdot 5

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 5) \cdot 5

規則を適用

- (- a) = a

= 1 + 4 \cdot 5 \cdot 5

数を乗じる:

4 \cdot 5 \cdot 5 = 100

= 1 + 100

数を足す:

1 + 100 = 101

= 101

u_{1, 2} = \frac{- 1 \pm 101}{2 ( - 5 )}

解を分離する

u_{1} = \frac{- 1 + 101}{2 ( - 5 )}, u_{2} = \frac{- 1 - 101}{2 ( - 5 )}

u = \frac{- 1 + 101}{2 ( - 5 )} : - \frac{- 1 + 101}{10}

\frac{- 1 + 101}{2 ( - 5 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 101}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{- 1 + 101}{- 10}

分数の規則を適用する:

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

= - \frac{- 1 + 101}{10}

u = \frac{- 1 - 101}{2 ( - 5 )} : \frac{1 + 101}{10}

\frac{- 1 - 101}{2 ( - 5 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 101}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{- 1 - 101}{- 10}

分数の規則を適用する:

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

- 1 - 101 = - (1 + 101)

= \frac{1 + 101}{10}

二次equationの解:

u = - \frac{- 1 + 101}{10}, u = \frac{1 + 101}{10}

代用を戻す

u = sin (x)

sin (x) = - \frac{- 1 + 101}{10}, sin (x) = \frac{1 + 101}{10}

sin (x) = - \frac{- 1 + 101}{10}, sin (x) = \frac{1 + 101}{10}

sin (x) = - \frac{- 1 + 101}{10} : x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

sin (x) = - \frac{- 1 + 101}{10}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{- 1 + 101}{10}

以下の一般解

sin (x) = - \frac{- 1 + 101}{10}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

sin (x) = \frac{1 + 101}{10} :

解なし

sin (x) = \frac{1 + 101}{10}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

5 cos^{2} (x) - sin (x) = 0 : x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

5 cos^{2} (x) - sin (x) = 0

三角関数の公式を使用して書き換える

- sin (x) + 5 cos^{2} (x)

ピタゴラスの公式を使用する:

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

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

= - sin (x) + 5 (1 - sin^{2} (x))

- sin (x) + (1 - sin^{2} (x)) \cdot 5 = 0

置換で解く

- sin (x) + (1 - sin^{2} (x)) \cdot 5 = 0

仮定:

sin (x) = u

- u + (1 - u^{2}) \cdot 5 = 0

- u + (1 - u^{2}) \cdot 5 = 0 : u = - \frac{1 + 101}{10}, u = \frac{101 - 1}{10}

- u + (1 - u^{2}) \cdot 5 = 0

拡張

- u + (1 - u^{2}) \cdot 5 : - u + 5 - 5 u^{2}

- u + (1 - u^{2}) \cdot 5

= - u + 5 (1 - u^{2})

拡張

5 (1 - u^{2}) : 5 - 5 u^{2}

5 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = 5, b = 1, c = u^{2}

= 5 \cdot 1 - 5 u^{2}

数を乗じる:

5 \cdot 1 = 5

= 5 - 5 u^{2}

= - u + 5 - 5 u^{2}

- u + 5 - 5 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- 5 u^{2} - u + 5 = 0

解くとthe二次式

- 5 u^{2} - u + 5 = 0

二次Equationの公式:

次の場合:

a = - 5, b = - 1, c = 5

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 5 ) \cdot 5}{2 ( - 5 )}

(- 1)^{2} - 4 (- 5) \cdot 5 = 101

(- 1)^{2} - 4 (- 5) \cdot 5

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 5 \cdot 5

(- 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 5 \cdot 5 = 100

4 \cdot 5 \cdot 5

数を乗じる:

4 \cdot 5 \cdot 5 = 100

= 100

= 1 + 100

数を足す:

1 + 100 = 101

= 101

u_{1, 2} = \frac{- ( - 1 ) \pm 101}{2 ( - 5 )}

解を分離する

u_{1} = \frac{- ( - 1 ) + 101}{2 ( - 5 )}, u_{2} = \frac{- ( - 1 ) - 101}{2 ( - 5 )}

u = \frac{- ( - 1 ) + 101}{2 ( - 5 )} : - \frac{1 + 101}{10}

\frac{- ( - 1 ) + 101}{2 ( - 5 )}

括弧を削除する:

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

= \frac{1 + 101}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{1 + 101}{- 10}

分数の規則を適用する:

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

= - \frac{1 + 101}{10}

u = \frac{- ( - 1 ) - 101}{2 ( - 5 )} : \frac{101 - 1}{10}

\frac{- ( - 1 ) - 101}{2 ( - 5 )}

括弧を削除する:

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

= \frac{1 - 101}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{1 - 101}{- 10}

分数の規則を適用する:

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

1 - 101 = - (101 - 1)

= \frac{101 - 1}{10}

二次equationの解:

u = - \frac{1 + 101}{10}, u = \frac{101 - 1}{10}

代用を戻す

u = sin (x)

sin (x) = - \frac{1 + 101}{10}, sin (x) = \frac{101 - 1}{10}

sin (x) = - \frac{1 + 101}{10}, sin (x) = \frac{101 - 1}{10}

sin (x) = - \frac{1 + 101}{10} :

解なし

sin (x) = - \frac{1 + 101}{10}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = \frac{101 - 1}{10} : x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

sin (x) = \frac{101 - 1}{10}

三角関数の逆数プロパティを適用する

sin (x) = \frac{101 - 1}{10}

以下の一般解

sin (x) = \frac{101 - 1}{10}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

すべての解を組み合わせる

x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

すべての解を組み合わせる

x = arcsin (- \frac{- 1 + 101}{10}) + 2 πn, x = π + arcsin (\frac{- 1 + 101}{10}) + 2 πn, x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

元のequationに当てはめて解を検算する

5 cos (x) cot (x) = 1

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

arcsin (- \frac{- 1 + 101}{10}) + 2 πn :

偽

arcsin (- \frac{- 1 + 101}{10}) + 2 πn

挿入

n = 1

arcsin (- \frac{- 1 + 101}{10}) + 2 π 1

5 cos (x) cot (x) = 1

の挿入向け

x = arcsin (- \frac{- 1 + 101}{10}) + 2 π 1

5 cos (arcsin (- \frac{- 1 + 101}{10}) + 2 π 1) cot (arcsin (- \frac{- 1 + 101}{10}) + 2 π 1) = 1

改良

- 1 = 1

\Rightarrow 偽

解答を確認する

π + arcsin (\frac{- 1 + 101}{10}) + 2 πn :

偽

π + arcsin (\frac{- 1 + 101}{10}) + 2 πn

挿入

n = 1

π + arcsin (\frac{- 1 + 101}{10}) + 2 π 1

5 cos (x) cot (x) = 1

の挿入向け

x = π + arcsin (\frac{- 1 + 101}{10}) + 2 π 1

5 cos (π + arcsin (\frac{- 1 + 101}{10}) + 2 π 1) cot (π + arcsin (\frac{- 1 + 101}{10}) + 2 π 1) = 1

改良

- 1 = 1

\Rightarrow 偽

解答を確認する

arcsin (\frac{101 - 1}{10}) + 2 πn :

真

arcsin (\frac{101 - 1}{10}) + 2 πn

挿入

n = 1

arcsin (\frac{101 - 1}{10}) + 2 π 1

5 cos (x) cot (x) = 1

の挿入向け

x = arcsin (\frac{101 - 1}{10}) + 2 π 1

5 cos (arcsin (\frac{101 - 1}{10}) + 2 π 1) cot (arcsin (\frac{101 - 1}{10}) + 2 π 1) = 1

改良

1 = 1

\Rightarrow 真

解答を確認する

π - arcsin (\frac{101 - 1}{10}) + 2 πn :

真

π - arcsin (\frac{101 - 1}{10}) + 2 πn

挿入

n = 1

π - arcsin (\frac{101 - 1}{10}) + 2 π 1

5 cos (x) cot (x) = 1

の挿入向け

x = π - arcsin (\frac{101 - 1}{10}) + 2 π 1

5 cos (π - arcsin (\frac{101 - 1}{10}) + 2 π 1) cot (π - arcsin (\frac{101 - 1}{10}) + 2 π 1) = 1

改良

1 = 1

\Rightarrow 真

x = arcsin (\frac{101 - 1}{10}) + 2 πn, x = π - arcsin (\frac{101 - 1}{10}) + 2 πn

10進法形式で解を証明する

x = 1.13135 \dots + 2 πn, x = π - 1.13135 \dots + 2 πn

グラフ

人気の例

- 0.5 = cos (3 x)

sin (3 x) = - 0.5

tan (x) = \frac{5}{8}

cos (x) = \frac{7}{8}

cos(6x)=1,0<= x<= pi/2

cos (6 x) = 1, 0 \leq x \leq \frac{π}{2}