ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある >

csc^2(x)=sec(x)

解

csc^{2} (x) = sec (x)

解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

csc^{2} (x) = sec (x)

両辺から

sec (x)

を引く

csc^{2} (x) - sec (x) = 0

サイン, コサインで表わす

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

簡素化

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

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

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

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

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

以下の最小公倍数:

sin^{2} (x), cos (x) : sin^{2} (x) cos (x)

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

最小公倍数 (LCM)

sin^{2} (x)

または以下のいずれかに現れる因数で構成された式を計算する:

cos (x)

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

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

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

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

の場合

:

分母と分子に以下を乗じる:

cos (x)

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

\frac{1}{cos ( x )}

の場合

:

分母と分子に以下を乗じる:

sin^{2} (x)

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

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

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

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

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

両辺に

sin^{2} (x)

を足す

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

両辺を2乗する

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

両辺から

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

を引く

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

因数

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

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

指数の規則を適用する:

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

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

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

2乗の差の公式を適用する:

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

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

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

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

各部分を別個に解く

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

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

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

三角関数の公式を使用して書き換える

cos (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)

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

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

置換で解く

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

仮定:

cos (x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

標準的な形式で書く

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

- u^{2} + u + 1 = 0

解くとthe二次式

- u^{2} + u + 1 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

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

解を分離する

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

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

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

括弧を削除する:

(- a) = - a

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

数を乗じる:

2 \cdot 1 = 2

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

分数の規則を適用する:

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

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

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

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

括弧を削除する:

(- a) = - a

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

数を乗じる:

2 \cdot 1 = 2

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

分数の規則を適用する:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次equationの解:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

代用を戻す

u = cos (x)

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2} : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{1 + 5}{2} :

解なし

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

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

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

三角関数の公式を使用して書き換える

cos (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)

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

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

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

括弧を分配する

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

マイナス・プラスの規則を適用する

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

標準的な形式で書く

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

u^{2} + u - 1 = 0

解くとthe二次式

u^{2} + u - 1 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

解を分離する

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

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

数を乗じる:

2 \cdot 1 = 2

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

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

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

数を乗じる:

2 \cdot 1 = 2

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

二次equationの解:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

代用を戻す

u = cos (x)

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2} : x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

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

解なし

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

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

すべての解を組み合わせる

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

元のequationに当てはめて解を検算する

csc^{2} (x) = sec (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

arccos (- \frac{- 1 + 5}{2}) + 2 πn :

偽

arccos (- \frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (x) = sec (x)

の挿入向け

x = arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = sec (arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

改良

1.61803 \dots = - 1.61803 \dots

\Rightarrow 偽

解答を確認する

- arccos (- \frac{- 1 + 5}{2}) + 2 πn :

偽

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

- arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (x) = sec (x)

の挿入向け

x = - arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = sec (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

改良

1.61803 \dots = - 1.61803 \dots

\Rightarrow 偽

解答を確認する

arccos (\frac{- 1 + 5}{2}) + 2 πn :

真

arccos (\frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (x) = sec (x)

の挿入向け

x = arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (arccos (\frac{- 1 + 5}{2}) + 2 π 1) = sec (arccos (\frac{- 1 + 5}{2}) + 2 π 1)

改良

1.61803 \dots = 1.61803 \dots

\Rightarrow 真

解答を確認する

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn :

真

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (x) = sec (x)

の挿入向け

x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1) = sec (2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1)

改良

1.61803 \dots = 1.61803 \dots

\Rightarrow 真

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

10進法形式で解を証明する

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

人気の例

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

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

6cos^3(x)+cos^2(x)-1=0

6 cos^{3} (x) + cos^{2} (x) - 1 = 0

4tan^2(x)+12tan(x)-27=0

4 tan^{2} (x) + 12 tan (x) - 27 = 0

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

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

cos^4(a)=3+4cos^2(a)+cos^4(a)

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)