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人気のある三角関数 >

sec(x)-(sin(x))/(cos(x))=cos(x)

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

cos (x)

を引く

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

簡素化

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

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

元を分数に変換する:

sec (x) = \frac{sec ( x ) cos ( x )}{cos ( x )}, cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}

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

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

sec (x) cos (x) - sin (x) - cos (x) cos (x)

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

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)

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

サイン, コサインで表わす

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

簡素化

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

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

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

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

分数を乗じる:

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

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

共通因数を約分する:

cos (x)

= 1

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

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

両辺に

cos^{2} (x)

を足す

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

両辺を2乗する

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

両辺から

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

を引く

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

因数

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

(1 - sin (x))^{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}

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

2乗の差の公式を適用する:

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

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

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

改良

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

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

各部分を別個に解く

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

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

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

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

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

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

簡素化

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

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

置換で解く

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

仮定:

sin (x) = u

2 - u - u^{2} = 0

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

2 - u - u^{2} = 0

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

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

(- 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 2 = 8

4 \cdot 1 \cdot 2

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

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

解を分離する

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

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

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

括弧を削除する:

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

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

数を足す:

1 + 3 = 4

= \frac{4}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{4}{- 2}

分数の規則を適用する:

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

= - \frac{4}{2}

数を割る:

\frac{4}{2} = 2

= - 2

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

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

括弧を削除する:

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

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

数を引く:

1 - 3 = - 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 = - 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

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

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

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

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

(- 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

数を引く:

1 - 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

規則を適用

- (- 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}

規則を適用

\frac{a}{a} = 1

= 1

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

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

規則を適用

- (- 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}

規則を適用

\frac{0}{a} = 0, a \neq = 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{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) = 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{π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

すべての解を組み合わせる

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

元のequationに当てはめて解を検算する

sec (x) - \frac{sin ( x )}{cos ( x )} = cos (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}{2} + 2 πn :

偽

\frac{π}{2} + 2 πn

挿入

n = 1

\frac{π}{2} + 2 π 1

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

の挿入向け

x = \frac{π}{2} + 2 π 1

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

未定義

\Rightarrow 偽

解答を確認する

2 πn :

真

2 πn

挿入

n = 1

2 π 1

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

の挿入向け

x = 2 π 1

sec (2 π 1) - \frac{sin ( 2 π 1 )}{cos ( 2 π 1 )} = cos (2 π 1)

改良

1 = 1

\Rightarrow 真

解答を確認する

π + 2 πn :

真

π + 2 πn

挿入

n = 1

π + 2 π 1

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

の挿入向け

x = π + 2 π 1

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

改良

- 1 = - 1

\Rightarrow 真

x = 2 πn, x = π + 2 πn

グラフ

人気の例

cos(θ)=0.4,sec(θ)

cos (θ) = 0.4, sec (θ)

cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ))

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

sec(3x)-csc(30)=0,(x+35)/5

sec (3 x) - csc (3 0^{\circ}) = 0, \frac{x + 35}{5}

4cos(2θ)-10cos(θ)+14=7,0<= θ<360

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

2 tan^{2} (θ) = 2