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人気のある三角関数 >

2cos(x)+3tan(x)=3sec(x)

解

2 cos (x) + 3 tan (x) = 3 sec (x)

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 cos (x) + 3 tan (x) = 3 sec (x)

両辺から

3 sec (x)

を引く

2 cos (x) + 3 tan (x) - 3 sec (x) = 0

サイン, コサインで表わす

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

簡素化

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

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

3 \cdot \frac{sin ( x )}{cos ( x )} = \frac{3 sin ( x )}{cos ( x )}

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

分数を乗じる:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

3 \cdot \frac{1}{cos ( x )} = \frac{3}{cos ( x )}

3 \cdot \frac{1}{cos ( x )}

分数を乗じる:

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

= \frac{1 \cdot 3}{cos ( x )}

数を乗じる:

1 \cdot 3 = 3

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

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

分数を組み合わせる

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

規則を適用

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

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

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

元を分数に変換する:

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

= \frac{2 cos ( x ) cos ( x )}{cos ( x )} + \frac{sin ( x ) \cdot 3 - 3}{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{2 cos ( x ) cos ( x ) + sin ( x ) \cdot 3 - 3}{cos ( x )}

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

2 cos (x) cos (x) + sin (x) \cdot 3 - 3

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

2 cos (x) 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)

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

数を足す:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

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

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

両辺から

3 sin (x)

を引く

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

両辺を2乗する

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

両辺から

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

を引く

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x) = 0

因数

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

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

を書き換え

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

(2 cos^{2} (x) - 3)^{2} - 9 sin^{2} (x)

9

を書き換え

3^{2}

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

指数の規則を適用する:

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

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

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

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

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

改良

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

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

各部分を別個に解く

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

2 cos^{2} (x) - 3 + 3 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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

三角関数の公式を使用して書き換える

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

数を足す/引く:

- 3 + 2 = - 1

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

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

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

置換で解く

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

仮定:

sin (x) = u

- 1 - 2 u^{2} + 3 u = 0

- 1 - 2 u^{2} + 3 u = 0 : u = \frac{1}{2}, u = 1

- 1 - 2 u^{2} + 3 u = 0

標準的な形式で書く

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

- 2 u^{2} + 3 u - 1 = 0

解くとthe二次式

- 2 u^{2} + 3 u - 1 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 3, c = - 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

3^{2} - 4 (- 2) (- 1) = 1

3^{2} - 4 (- 2) (- 1)

規則を適用

- (- a) = a

= 3^{2} - 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数を引く:

9 - 8 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 3 + 1}{2 ( - 2 )}, u_{2} = \frac{- 3 - 1}{2 ( - 2 )}

u = \frac{- 3 + 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 3 + 1}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 2}

数を足す/引く:

- 3 + 1 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 3 - 1}{2 ( - 2 )} : 1

\frac{- 3 - 1}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 2}

数を引く:

- 3 - 1 = - 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 = \frac{1}{2}, u = 1

代用を戻す

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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

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

三角関数の公式を使用して書き換える

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

数を足す/引く:

- 3 + 2 = - 1

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

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

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

置換で解く

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

仮定:

sin (x) = u

- 1 - 2 u^{2} - 3 u = 0

- 1 - 2 u^{2} - 3 u = 0 : u = - 1, u = - \frac{1}{2}

- 1 - 2 u^{2} - 3 u = 0

標準的な形式で書く

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

- 2 u^{2} - 3 u - 1 = 0

解くとthe二次式

- 2 u^{2} - 3 u - 1 = 0

二次Equationの公式:

次の場合:

a = - 2, b = - 3, c = - 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) (- 1) = 1

(- 3)^{2} - 4 (- 2) (- 1)

規則を適用

- (- a) = a

= (- 3)^{2} - 4 \cdot 2 \cdot 1

指数の規則を適用する:

n

が偶数であれば

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

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

= 3^{2} - 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数を引く:

9 - 8 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- ( - 3 ) + 1}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 1}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 1}{2 ( - 2 )} : - 1

\frac{- ( - 3 ) + 1}{2 ( - 2 )}

括弧を削除する:

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

= \frac{3 + 1}{- 2 \cdot 2}

数を足す:

3 + 1 = 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

u = \frac{- ( - 3 ) - 1}{2 ( - 2 )} : - \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 ( - 2 )}

括弧を削除する:

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

= \frac{3 - 1}{- 2 \cdot 2}

数を引く:

3 - 1 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{- 4}

分数の規則を適用する:

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

= - \frac{2}{4}

共通因数を約分する:

2

= - \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{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) = - \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{3 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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 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

2 cos (x) + 3 tan (x) = 3 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

\frac{π}{6} + 2 πn :

真

\frac{π}{6} + 2 πn

挿入

n = 1

\frac{π}{6} + 2 π 1

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{π}{6} + 2 π 1

2 cos (\frac{π}{6} + 2 π 1) + 3 tan (\frac{π}{6} + 2 π 1) = 3 sec (\frac{π}{6} + 2 π 1)

改良

3.46410 \dots = 3.46410 \dots

\Rightarrow 真

解答を確認する

\frac{5 π}{6} + 2 πn :

真

\frac{5 π}{6} + 2 πn

挿入

n = 1

\frac{5 π}{6} + 2 π 1

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{5 π}{6} + 2 π 1

2 cos (\frac{5 π}{6} + 2 π 1) + 3 tan (\frac{5 π}{6} + 2 π 1) = 3 sec (\frac{5 π}{6} + 2 π 1)

改良

- 3.46410 \dots = - 3.46410 \dots

\Rightarrow 真

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

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{π}{2} + 2 π 1

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

改良

\infty = \infty

\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

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{3 π}{2} + 2 π 1

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

改良

\infty = - \infty

\Rightarrow 偽

解答を確認する

\frac{7 π}{6} + 2 πn :

偽

\frac{7 π}{6} + 2 πn

挿入

n = 1

\frac{7 π}{6} + 2 π 1

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{7 π}{6} + 2 π 1

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

改良

0 = - 3.46410 \dots

\Rightarrow 偽

解答を確認する

\frac{11 π}{6} + 2 πn :

偽

\frac{11 π}{6} + 2 πn

挿入

n = 1

\frac{11 π}{6} + 2 π 1

2 cos (x) + 3 tan (x) = 3 sec (x)

の挿入向け

x = \frac{11 π}{6} + 2 π 1

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

改良

0 = 3.46410 \dots

\Rightarrow 偽

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

equationは以下で未定義のため:

\frac{π}{2} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

グラフ

人気の例

14tan^2(x)=-14tan(x)

14 tan^{2} (x) = - 14 tan (x)

cos (3 x - 1) = 0

tan(x+7)=cot(4x+8)

tan (x + 7) = cot (4 x + 8)

4sqrt(3)cos(t)-4sin(t)=0

43 cos (t) - 4 sin (t) = 0

cos(x)=(4.1)/(7.9)

cos (x) = \frac{4.1}{7.9}