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人気のある三角関数 >

(tan(x)-sec(x))^2=3

解

(tan (x) - sec (x))^{2} = 3

解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

解答ステップ

(tan (x) - sec (x))^{2} = 3

両辺から

3

を引く

(tan (x) - sec (x))^{2} - 3 = 0

サイン, コサインで表わす

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3 = 0

簡素化

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3 : \frac{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3

分数を組み合わせる

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

規則を適用

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

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

= (\frac{sin ( x ) - 1}{cos ( x )})^{2} - 3

指数の規則を適用する:

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

= \frac{( sin ( x ) - 1 ) ^{2}}{cos ^{2} ( x )} - 3

元を分数に変換する:

3 = \frac{3 cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{( sin ( x ) - 1 ) ^{2}}{cos ^{2} ( x )} - \frac{3 cos ^{2} ( x )}{cos ^{2} ( 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 ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )} = 0

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

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

両辺に

3 cos^{2} (x)

を足す

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

両辺を2乗する

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

両辺から

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

を引く

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) = 0

因数

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) : (sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x)) (sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x))

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

を書き換え

(sin^{2} (x) - 2 sin (x) + 1)^{2} - (3 cos^{2} (x))^{2}

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

9

を書き換え

3^{2}

= (sin^{2} (x) - 2 sin (x) + 1)^{2} - 3^{2} cos^{4} (x)

指数の規則を適用する:

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

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

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

指数の規則を適用する:

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

3^{2} (cos^{2} (x))^{2} = (3 cos^{2} (x))^{2}

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

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

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

改良

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

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

各部分を別個に解く

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

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

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

三角関数の公式を使用して書き換える

1 + sin^{2} (x) - 2 sin (x) + 3 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)

= 1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x))

簡素化

1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x)) : - 2 sin^{2} (x) - 2 sin (x) + 4

1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x))

拡張

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

数を乗じる:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (x)

= 1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x)

簡素化

1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x) : - 2 sin^{2} (x) - 2 sin (x) + 4

1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x)

条件のようなグループ

= sin^{2} (x) - 2 sin (x) - 3 sin^{2} (x) + 1 + 3

類似した元を足す:

sin^{2} (x) - 3 sin^{2} (x) = - 2 sin^{2} (x)

= - 2 sin^{2} (x) - 2 sin (x) + 1 + 3

数を足す:

1 + 3 = 4

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

4 - 2 u - 2 u^{2} = 0 : u = - 2, u = 1

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = - 2, c = 4

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

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

(- 2)^{2} - 4 (- 2) \cdot 4 = 6

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

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

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

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

数を乗じる:

4 \cdot 2 \cdot 4 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- ( - 2 ) + 6}{2 ( - 2 )}, u_{2} = \frac{- ( - 2 ) - 6}{2 ( - 2 )}

u = \frac{- ( - 2 ) + 6}{2 ( - 2 )} : - 2

\frac{- ( - 2 ) + 6}{2 ( - 2 )}

括弧を削除する:

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

= \frac{2 + 6}{- 2 \cdot 2}

数を足す:

2 + 6 = 8

= \frac{8}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{8}{- 4}

分数の規則を適用する:

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

= - \frac{8}{4}

数を割る:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 2 ) - 6}{2 ( - 2 )} : 1

\frac{- ( - 2 ) - 6}{2 ( - 2 )}

括弧を削除する:

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

= \frac{2 - 6}{- 2 \cdot 2}

数を引く:

2 - 6 = - 4

= \frac{- 4}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

分数の規則を適用する:

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

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

u = - 2, u = 1

代用を戻す

u = sin (x)

sin (x) = - 2, sin (x) = 1

sin (x) = - 2, sin (x) = 1

sin (x) = - 2 :

解なし

sin (x) = - 2

- 1 \leq sin (x) \leq 1

解なし

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 = \frac{π}{2} + 2 πn

sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

三角関数の公式を使用して書き換える

1 + sin^{2} (x) - 2 sin (x) - 3 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)

= 1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x))

簡素化

1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x)) : 4 sin^{2} (x) - 2 sin (x) - 2

1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x))

拡張

- 3 (1 - sin^{2} (x)) : - 3 + 3 sin^{2} (x)

- 3 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (x)

= - 3 \cdot 1 - (- 3) sin^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (x)

数を乗じる:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (x)

= 1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x)

簡素化

1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x) : 4 sin^{2} (x) - 2 sin (x) - 2

1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x)

条件のようなグループ

= sin^{2} (x) - 2 sin (x) + 3 sin^{2} (x) + 1 - 3

類似した元を足す:

sin^{2} (x) + 3 sin^{2} (x) = 4 sin^{2} (x)

= 4 sin^{2} (x) - 2 sin (x) + 1 - 3

数を足す/引く:

1 - 3 = - 2

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

- 2 - 2 u + 4 u^{2} = 0 : u = 1, u = - \frac{1}{2}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 4, b = - 2, c = - 2

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

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

(- 2)^{2} - 4 \cdot 4 (- 2) = 6

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

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

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

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

数を乗じる:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 4}

解を分離する

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 4}

u = \frac{- ( - 2 ) + 6}{2 \cdot 4} : 1

\frac{- ( - 2 ) + 6}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{2 + 6}{2 \cdot 4}

数を足す:

2 + 6 = 8

= \frac{8}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 6}{2 \cdot 4} : - \frac{1}{2}

\frac{- ( - 2 ) - 6}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{2 - 6}{2 \cdot 4}

数を引く:

2 - 6 = - 4

= \frac{- 4}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 4}{8}

分数の規則を適用する:

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

= - \frac{4}{8}

共通因数を約分する:

4

= - \frac{1}{2}

二次equationの解:

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

代用を戻す

u = sin (x)

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

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

以下の一般解

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

元のequationに当てはめて解を検算する

(tan (x) - sec (x))^{2} = 3

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

\frac{π}{2} + 2 πn :

偽

\frac{π}{2} + 2 πn

挿入

n = 1

\frac{π}{2} + 2 π 1

(tan (x) - sec (x))^{2} = 3

の挿入向け

x = \frac{π}{2} + 2 π 1

(tan (\frac{π}{2} + 2 π 1) - sec (\frac{π}{2} + 2 π 1))^{2} = 3

未定義

\Rightarrow 偽

解答を確認する

\frac{7 π}{6} + 2 πn :

真

\frac{7 π}{6} + 2 πn

挿入

n = 1

\frac{7 π}{6} + 2 π 1

(tan (x) - sec (x))^{2} = 3

の挿入向け

x = \frac{7 π}{6} + 2 π 1

(tan (\frac{7 π}{6} + 2 π 1) - sec (\frac{7 π}{6} + 2 π 1))^{2} = 3

改良

3 = 3

\Rightarrow 真

解答を確認する

\frac{11 π}{6} + 2 πn :

真

\frac{11 π}{6} + 2 πn

挿入

n = 1

\frac{11 π}{6} + 2 π 1

(tan (x) - sec (x))^{2} = 3

の挿入向け

x = \frac{11 π}{6} + 2 π 1

(tan (\frac{11 π}{6} + 2 π 1) - sec (\frac{11 π}{6} + 2 π 1))^{2} = 3

改良

3 = 3

\Rightarrow 真

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

グラフ

人気の例

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