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人気のある三角関数 >

1/(1-cos(x))+1/(1+cos(x))=2csc(x)

解

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

解

x = \frac{π}{2} + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

2 csc (x)

を引く

\frac{2 cos ^{2} ( x ) csc ( x ) - 2 csc ( x ) + 2}{( - cos ( x ) + 1 ) ( cos ( x ) + 1 )} = 0

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

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

サイン, コサインで表わす

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

簡素化

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

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

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

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

分数を乗じる:

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

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

数を乗じる:

1 \cdot 2 = 2

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

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

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

分数を乗じる:

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

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

数を乗じる:

1 \cdot 2 = 2

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

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

分数を組み合わせる

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

規則を適用

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

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

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

元を分数に変換する:

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

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

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

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

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

両辺から

2 sin (x)

を引く

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

両辺を2乗する

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

両辺から

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

を引く

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

因数

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

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

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

を書き換え

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

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

4

を書き換え

2^{2}

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

指数の規則を適用する:

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

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

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

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

2乗の差の公式を適用する:

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

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

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

改良

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

因数

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

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

共通項をくくり出す

2

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

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

因数

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

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

共通項をくくり出す

2

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

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

改良

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

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

各部分を別個に解く

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

標準的な形式で書く

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

- u^{2} + u = 0

解くとthe二次式

- u^{2} + u = 0

二次Equationの公式:

次の場合:

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

規則を適用

0 \cdot a = 0

= 1 + 0

数を足す:

1 + 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

括弧を削除する:

(- a) = - a

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

数を足す/引く:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{0}{- 2}

分数の規則を適用する:

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

= - \frac{0}{2}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

括弧を削除する:

(- a) = - a

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

数を引く:

- 1 - 1 = - 2

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

数を乗じる:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

分数の規則を適用する:

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

= \frac{2}{2}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

u = 0, u = 1

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

以下の一般解

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

すべての解を組み合わせる

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

cos^{2} (x) - 1 - sin (x) = 0 : x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

- u - u^{2} = 0

- u - u^{2} = 0 : u = - 1, u = 0

- u - u^{2} = 0

標準的な形式で書く

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

- u^{2} - u = 0

解くとthe二次式

- u^{2} - u = 0

二次Equationの公式:

次の場合:

a = - 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 0 = 1

(- 1)^{2} - 4 (- 1) \cdot 0

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 0

(- 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 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

規則を適用

0 \cdot a = 0

= 0

= 1 + 0

数を足す:

1 + 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

括弧を削除する:

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

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

数を足す:

1 + 1 = 2

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

数を乗じる:

2 \cdot 1 = 2

= \frac{2}{- 2}

分数の規則を適用する:

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

= - \frac{2}{2}

規則を適用

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 1}{2 ( - 1 )} : 0

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

括弧を削除する:

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

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

数を引く:

1 - 1 = 0

= \frac{0}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{0}{- 2}

分数の規則を適用する:

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

= - \frac{0}{2}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

二次equationの解:

u = - 1, u = 0

代用を戻す

u = sin (x)

sin (x) = - 1, sin (x) = 0

sin (x) = - 1, sin (x) = 0

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

sin (x) = - 1

以下の一般解

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

すべての解を組み合わせる

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

すべての解を組み合わせる

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

元のequationに当てはめて解を検算する

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

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

2 πn :

偽

2 πn

挿入

n = 1

2 π 1

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

の挿入向け

x = 2 π 1

\frac{1}{1 - cos ( 2 π 1 )} + \frac{1}{1 + cos ( 2 π 1 )} = 2 csc (2 π 1)

未定義

\Rightarrow 偽

解答を確認する

π + 2 πn :

偽

π + 2 πn

挿入

n = 1

π + 2 π 1

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

の挿入向け

x = π + 2 π 1

\frac{1}{1 - cos ( π + 2 π 1 )} + \frac{1}{1 + cos ( π + 2 π 1 )} = 2 csc (π + 2 π 1)

未定義

\Rightarrow 偽

解答を確認する

\frac{π}{2} + 2 πn :

真

\frac{π}{2} + 2 πn

挿入

n = 1

\frac{π}{2} + 2 π 1

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

の挿入向け

x = \frac{π}{2} + 2 π 1

\frac{1}{1 - cos ( \frac{π}{2} + 2 π 1 )} + \frac{1}{1 + cos ( \frac{π}{2} + 2 π 1 )} = 2 csc (\frac{π}{2} + 2 π 1)

改良

2 = 2

\Rightarrow 真

解答を確認する

\frac{3 π}{2} + 2 πn :

偽

\frac{3 π}{2} + 2 πn

挿入

n = 1

\frac{3 π}{2} + 2 π 1

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

の挿入向け

x = \frac{3 π}{2} + 2 π 1

\frac{1}{1 - cos ( \frac{3 π}{2} + 2 π 1 )} + \frac{1}{1 + cos ( \frac{3 π}{2} + 2 π 1 )} = 2 csc (\frac{3 π}{2} + 2 π 1)

改良

2 = - 2

\Rightarrow 偽

x = \frac{π}{2} + 2 πn

グラフ

人気の例

cos (θ) = 0.75

4.58^2=3.3^2+3.3^2-2(3.3)(3.3)cos(x)

4.5 8^{2} = 3. 3^{2} + 3. 3^{2} - 2 (3.3) (3.3) cos (x)

4 - 3 sin (θ) = 7

sin(x)cos(x)=-1/4 ,0<= x<= 2pi

sin (x) cos (x) = - \frac{1}{4}, 0 \leq x \leq 2 π

tan (x) = (\frac{4}{3})