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人気のある >

sin(x)tan(x)+sec(x)+5=0

解

sin (x) tan (x) + sec (x) + 5 = 0

解

x = 1.95226 \dots + 2 πn, x = - 1.95226 \dots + 2 πn

+1

度

x = 111.85638 \dots^{\circ} + 36 0^{\circ} n, x = - 111.85638 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sin (x) tan (x) + sec (x) + 5 = 0

サイン, コサインで表わす

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

簡素化

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

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

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

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

分数を乗じる:

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

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

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

sin (x) sin (x)

指数の規則を適用する:

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

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

= sin^{1 + 1} (x)

数を足す:

1 + 1 = 2

= sin^{2} (x)

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

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

分数を組み合わせる

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

規則を適用

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

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

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

元を分数に変換する:

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

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

\frac{sin ^{2} ( x ) + 1 + 5 cos ( x )}{cos ( x )} = 0

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

sin^{2} (x) + 1 + 5 cos (x) = 0

両辺から

5 cos (x)

を引く

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

両辺を2乗する

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

両辺から

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

を引く

(sin^{2} (x) + 1)^{2} - 25 cos^{2} (x) = 0

因数

(sin^{2} (x) + 1)^{2} - 25 cos^{2} (x) : (sin^{2} (x) + 1 + 5 cos (x)) (sin^{2} (x) + 1 - 5 cos (x))

(sin^{2} (x) + 1)^{2} - 25 cos^{2} (x)

(sin^{2} (x) + 1)^{2} - 25 cos^{2} (x)

を書き換え

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

(sin^{2} (x) + 1)^{2} - 25 cos^{2} (x)

25

を書き換え

5^{2}

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

指数の規則を適用する:

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

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

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

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

2乗の差の公式を適用する:

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

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

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

改良

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

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

各部分を別個に解く

sin^{2} (x) + 1 + 5 cos (x) = 0 or sin^{2} (x) + 1 - 5 cos (x) = 0

sin^{2} (x) + 1 + 5 cos (x) = 0 : x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

sin^{2} (x) + 1 + 5 cos (x) = 0

三角関数の公式を使用して書き換える

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

簡素化

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

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

置換で解く

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

仮定:

cos (x) = u

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

2 - u^{2} + 5 u = 0 : u = - \frac{- 5 + 33}{2}, u = \frac{5 + 33}{2}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 1, b = 5, c = 2

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

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

5^{2} - 4 (- 1) \cdot 2 = 33

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

規則を適用

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 2

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 5^{2} + 8

5^{2} = 25

= 25 + 8

数を足す:

25 + 8 = 33

= 33

u_{1, 2} = \frac{- 5 \pm 33}{2 ( - 1 )}

解を分離する

u_{1} = \frac{- 5 + 33}{2 ( - 1 )}, u_{2} = \frac{- 5 - 33}{2 ( - 1 )}

u = \frac{- 5 + 33}{2 ( - 1 )} : - \frac{- 5 + 33}{2}

\frac{- 5 + 33}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 + 33}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 5 + 33}{- 2}

分数の規則を適用する:

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

= - \frac{- 5 + 33}{2}

u = \frac{- 5 - 33}{2 ( - 1 )} : \frac{5 + 33}{2}

\frac{- 5 - 33}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 - 33}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 5 - 33}{- 2}

分数の規則を適用する:

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

- 5 - 33 = - (5 + 33)

= \frac{5 + 33}{2}

二次equationの解:

u = - \frac{- 5 + 33}{2}, u = \frac{5 + 33}{2}

代用を戻す

u = cos (x)

cos (x) = - \frac{- 5 + 33}{2}, cos (x) = \frac{5 + 33}{2}

cos (x) = - \frac{- 5 + 33}{2}, cos (x) = \frac{5 + 33}{2}

cos (x) = - \frac{- 5 + 33}{2} : x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

cos (x) = - \frac{- 5 + 33}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{- 5 + 33}{2}

以下の一般解

cos (x) = - \frac{- 5 + 33}{2}

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

x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

cos (x) = \frac{5 + 33}{2} :

解なし

cos (x) = \frac{5 + 33}{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{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

sin^{2} (x) + 1 - 5 cos (x) = 0 : x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

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

三角関数の公式を使用して書き換える

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

簡素化

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

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

置換で解く

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

仮定:

cos (x) = u

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

2 - u^{2} - 5 u = 0 : u = - \frac{5 + 33}{2}, u = \frac{33 - 5}{2}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 1, b = - 5, c = 2

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

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

(- 5)^{2} - 4 (- 1) \cdot 2 = 33

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

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

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 1 \cdot 2

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 5^{2} + 8

5^{2} = 25

= 25 + 8

数を足す:

25 + 8 = 33

= 33

u_{1, 2} = \frac{- ( - 5 ) \pm 33}{2 ( - 1 )}

解を分離する

u_{1} = \frac{- ( - 5 ) + 33}{2 ( - 1 )}, u_{2} = \frac{- ( - 5 ) - 33}{2 ( - 1 )}

u = \frac{- ( - 5 ) + 33}{2 ( - 1 )} : - \frac{5 + 33}{2}

\frac{- ( - 5 ) + 33}{2 ( - 1 )}

括弧を削除する:

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

= \frac{5 + 33}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{5 + 33}{- 2}

分数の規則を適用する:

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

= - \frac{5 + 33}{2}

u = \frac{- ( - 5 ) - 33}{2 ( - 1 )} : \frac{33 - 5}{2}

\frac{- ( - 5 ) - 33}{2 ( - 1 )}

括弧を削除する:

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

= \frac{5 - 33}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{5 - 33}{- 2}

分数の規則を適用する:

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

5 - 33 = - (33 - 5)

= \frac{33 - 5}{2}

二次equationの解:

u = - \frac{5 + 33}{2}, u = \frac{33 - 5}{2}

代用を戻す

u = cos (x)

cos (x) = - \frac{5 + 33}{2}, cos (x) = \frac{33 - 5}{2}

cos (x) = - \frac{5 + 33}{2}, cos (x) = \frac{33 - 5}{2}

cos (x) = - \frac{5 + 33}{2} :

解なし

cos (x) = - \frac{5 + 33}{2}

- 1 \leq cos (x) \leq 1

解なし

cos (x) = \frac{33 - 5}{2} : x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

cos (x) = \frac{33 - 5}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{33 - 5}{2}

以下の一般解

cos (x) = \frac{33 - 5}{2}

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

x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

すべての解を組み合わせる

x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = arccos (\frac{33 - 5}{2}) + 2 πn, x = 2 π - arccos (\frac{33 - 5}{2}) + 2 πn

元のequationに当てはめて解を検算する

sin (x) tan (x) + sec (x) + 5 = 0

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{- 5 + 33}{2}) + 2 πn :

真

arccos (- \frac{- 5 + 33}{2}) + 2 πn

挿入

n = 1

arccos (- \frac{- 5 + 33}{2}) + 2 π 1

sin (x) tan (x) + sec (x) + 5 = 0

の挿入向け

x = arccos (- \frac{- 5 + 33}{2}) + 2 π 1

sin (arccos (- \frac{- 5 + 33}{2}) + 2 π 1) tan (arccos (- \frac{- 5 + 33}{2}) + 2 π 1) + sec (arccos (- \frac{- 5 + 33}{2}) + 2 π 1) + 5 = 0

改良

0 = 0

\Rightarrow 真

解答を確認する

- arccos (- \frac{- 5 + 33}{2}) + 2 πn :

真

- arccos (- \frac{- 5 + 33}{2}) + 2 πn

挿入

n = 1

- arccos (- \frac{- 5 + 33}{2}) + 2 π 1

sin (x) tan (x) + sec (x) + 5 = 0

の挿入向け

x = - arccos (- \frac{- 5 + 33}{2}) + 2 π 1

sin (- arccos (- \frac{- 5 + 33}{2}) + 2 π 1) tan (- arccos (- \frac{- 5 + 33}{2}) + 2 π 1) + sec (- arccos (- \frac{- 5 + 33}{2}) + 2 π 1) + 5 = 0

改良

0 = 0

\Rightarrow 真

解答を確認する

arccos (\frac{33 - 5}{2}) + 2 πn :

偽

arccos (\frac{33 - 5}{2}) + 2 πn

挿入

n = 1

arccos (\frac{33 - 5}{2}) + 2 π 1

sin (x) tan (x) + sec (x) + 5 = 0

の挿入向け

x = arccos (\frac{33 - 5}{2}) + 2 π 1

sin (arccos (\frac{33 - 5}{2}) + 2 π 1) tan (arccos (\frac{33 - 5}{2}) + 2 π 1) + sec (arccos (\frac{33 - 5}{2}) + 2 π 1) + 5 = 0

改良

10 = 0

\Rightarrow 偽

解答を確認する

2 π - arccos (\frac{33 - 5}{2}) + 2 πn :

偽

2 π - arccos (\frac{33 - 5}{2}) + 2 πn

挿入

n = 1

2 π - arccos (\frac{33 - 5}{2}) + 2 π 1

sin (x) tan (x) + sec (x) + 5 = 0

の挿入向け

x = 2 π - arccos (\frac{33 - 5}{2}) + 2 π 1

sin (2 π - arccos (\frac{33 - 5}{2}) + 2 π 1) tan (2 π - arccos (\frac{33 - 5}{2}) + 2 π 1) + sec (2 π - arccos (\frac{33 - 5}{2}) + 2 π 1) + 5 = 0

改良

10 = 0

\Rightarrow 偽

x = arccos (- \frac{- 5 + 33}{2}) + 2 πn, x = - arccos (- \frac{- 5 + 33}{2}) + 2 πn

10進法形式で解を証明する

x = 1.95226 \dots + 2 πn, x = - 1.95226 \dots + 2 πn

グラフ

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