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人気のある三角関数 >

tan(x)=(5+cos(x))/(6sin(x)cos(x))

解

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

両辺から

\frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

を引く

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )} = 0

簡素化

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )} : \frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )}

tan (x) - \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

元を分数に変換する:

tan (x) = \frac{tan ( x ) 6 sin ( x ) cos ( x )}{6 sin ( x ) cos ( x )}

= \frac{tan ( x ) \cdot 6 sin ( x ) cos ( x )}{6 sin ( x ) cos ( x )} - \frac{5 + cos ( x )}{6 sin ( 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{tan ( x ) \cdot 6 sin ( x ) cos ( x ) - ( 5 + cos ( x ) )}{6 sin ( x ) cos ( x )}

拡張

tan (x) \cdot 6 sin (x) cos (x) - (5 + cos (x)) : tan (x) \cdot 6 sin (x) cos (x) - 5 - cos (x)

tan (x) \cdot 6 sin (x) cos (x) - (5 + cos (x))

= 6 tan (x) sin (x) cos (x) - (5 + cos (x))

- (5 + cos (x)) : - 5 - cos (x)

- (5 + cos (x))

括弧を分配する

= - (5) - (cos (x))

マイナス・プラスの規則を適用する

+ (- a) = - a

= - 5 - cos (x)

= tan (x) \cdot 6 sin (x) cos (x) - 5 - cos (x)

= \frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )}

\frac{6 tan ( x ) sin ( x ) cos ( x ) - 5 - cos ( x )}{6 sin ( x ) cos ( x )} = 0

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

6 tan (x) sin (x) cos (x) - 5 - cos (x) = 0

サイン, コサインで表わす

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x) = 0

簡素化

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x) : 6 sin^{2} (x) - 5 - cos (x)

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) - 5 - cos (x)

6 \cdot \frac{sin ( x )}{cos ( x )} sin (x) cos (x) = 6 sin^{2} (x)

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

分数を乗じる:

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

= \frac{sin ( x ) \cdot 6 sin ( x ) cos ( x )}{cos ( x )}

共通因数を約分する:

cos (x)

= sin (x) \cdot 6 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)

= 6 sin^{1 + 1} (x)

数を足す:

1 + 1 = 2

= 6 sin^{2} (x)

= 6 sin^{2} (x) - 5 - cos (x)

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

両辺に

cos (x)

を足す

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

両辺を2乗する

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

両辺から

cos^{2} (x)

を引く

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

因数

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x) : (6 sin^{2} (x) - 5 + cos (x)) (6 sin^{2} (x) - 5 - cos (x))

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

2乗の差の公式を適用する:

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

(6 sin^{2} (x) - 5)^{2} - cos^{2} (x) = ((6 sin^{2} (x) - 5) + cos (x)) ((6 sin^{2} (x) - 5) - cos (x))

= ((6 sin^{2} (x) - 5) + cos (x)) ((6 sin^{2} (x) - 5) - cos (x))

改良

= (6 sin^{2} (x) + cos (x) - 5) (6 sin^{2} (x) - cos (x) - 5)

(6 sin^{2} (x) - 5 + cos (x)) (6 sin^{2} (x) - 5 - cos (x)) = 0

各部分を別個に解く

6 sin^{2} (x) - 5 + cos (x) = 0 or 6 sin^{2} (x) - 5 - cos (x) = 0

6 sin^{2} (x) - 5 + cos (x) = 0 : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

三角関数の公式を使用して書き換える

- 5 + cos (x) + 6 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)

= - 5 + cos (x) + 6 (1 - cos^{2} (x))

簡素化

- 5 + cos (x) + 6 (1 - cos^{2} (x)) : cos (x) - 6 cos^{2} (x) + 1

- 5 + cos (x) + 6 (1 - cos^{2} (x))

拡張

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数を乗じる:

6 \cdot 1 = 6

= 6 - 6 cos^{2} (x)

= - 5 + cos (x) + 6 - 6 cos^{2} (x)

簡素化

- 5 + cos (x) + 6 - 6 cos^{2} (x) : cos (x) - 6 cos^{2} (x) + 1

- 5 + cos (x) + 6 - 6 cos^{2} (x)

条件のようなグループ

= cos (x) - 6 cos^{2} (x) - 5 + 6

数を足す/引く:

- 5 + 6 = 1

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

1 + u - 6 u^{2} = 0

1 + u - 6 u^{2} = 0 : u = - \frac{1}{3}, u = \frac{1}{2}

1 + u - 6 u^{2} = 0

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 6, b = 1, c = 1

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

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

1^{2} - 4 (- 6) \cdot 1 = 5

1^{2} - 4 (- 6) \cdot 1

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

数を乗じる:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

数を足す:

1 + 24 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 6 )}

解を分離する

u_{1} = \frac{- 1 + 5}{2 ( - 6 )}, u_{2} = \frac{- 1 - 5}{2 ( - 6 )}

u = \frac{- 1 + 5}{2 ( - 6 )} : - \frac{1}{3}

\frac{- 1 + 5}{2 ( - 6 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 6}

数を足す/引く:

- 1 + 5 = 4

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

数を乗じる:

2 \cdot 6 = 12

= \frac{4}{- 12}

分数の規則を適用する:

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

= - \frac{4}{12}

共通因数を約分する:

4

= - \frac{1}{3}

u = \frac{- 1 - 5}{2 ( - 6 )} : \frac{1}{2}

\frac{- 1 - 5}{2 ( - 6 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 6}

数を引く:

- 1 - 5 = - 6

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

数を乗じる:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

分数の規則を適用する:

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

= \frac{6}{12}

共通因数を約分する:

6

= \frac{1}{2}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = - \frac{1}{3} : x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

6 sin^{2} (x) - 5 - cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

三角関数の公式を使用して書き換える

- 5 - cos (x) + 6 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)

= - 5 - cos (x) + 6 (1 - cos^{2} (x))

簡素化

- 5 - cos (x) + 6 (1 - cos^{2} (x)) : - 6 cos^{2} (x) - cos (x) + 1

- 5 - cos (x) + 6 (1 - cos^{2} (x))

拡張

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数を乗じる:

6 \cdot 1 = 6

= 6 - 6 cos^{2} (x)

= - 5 - cos (x) + 6 - 6 cos^{2} (x)

簡素化

- 5 - cos (x) + 6 - 6 cos^{2} (x) : - 6 cos^{2} (x) - cos (x) + 1

- 5 - cos (x) + 6 - 6 cos^{2} (x)

条件のようなグループ

= - cos (x) - 6 cos^{2} (x) - 5 + 6

数を足す/引く:

- 5 + 6 = 1

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

1 - u - 6 u^{2} = 0

1 - u - 6 u^{2} = 0 : u = - \frac{1}{2}, u = \frac{1}{3}

1 - u - 6 u^{2} = 0

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 6, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 6) \cdot 1 = 5

(- 1)^{2} - 4 (- 6) \cdot 1

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 1

(- 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 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

数を乗じる:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

数を足す:

1 + 24 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 6 )}

解を分離する

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 6 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 6 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 6 )} : - \frac{1}{2}

\frac{- ( - 1 ) + 5}{2 ( - 6 )}

括弧を削除する:

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

= \frac{1 + 5}{- 2 \cdot 6}

数を足す:

1 + 5 = 6

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

数を乗じる:

2 \cdot 6 = 12

= \frac{6}{- 12}

分数の規則を適用する:

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

= - \frac{6}{12}

共通因数を約分する:

6

= - \frac{1}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 6 )} : \frac{1}{3}

\frac{- ( - 1 ) - 5}{2 ( - 6 )}

括弧を削除する:

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

= \frac{1 - 5}{- 2 \cdot 6}

数を引く:

1 - 5 = - 4

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

数を乗じる:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

分数の規則を適用する:

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

= \frac{4}{12}

共通因数を約分する:

4

= \frac{1}{3}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

以下の一般解

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

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

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

すべての解を組み合わせる

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

すべての解を組み合わせる

x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

元のequationに当てはめて解を検算する

tan (x) = \frac{5 + cos ( x )}{6 sin ( 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

arccos (- \frac{1}{3}) + 2 πn :

偽

arccos (- \frac{1}{3}) + 2 πn

挿入

n = 1

arccos (- \frac{1}{3}) + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = arccos (- \frac{1}{3}) + 2 π 1

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

改良

- 2.82842 \dots = - 2.47487 \dots

\Rightarrow 偽

解答を確認する

- arccos (- \frac{1}{3}) + 2 πn :

偽

- arccos (- \frac{1}{3}) + 2 πn

挿入

n = 1

- arccos (- \frac{1}{3}) + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = - arccos (- \frac{1}{3}) + 2 π 1

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

改良

2.82842 \dots = 2.47487 \dots

\Rightarrow 偽

解答を確認する

\frac{π}{3} + 2 πn :

偽

\frac{π}{3} + 2 πn

挿入

n = 1

\frac{π}{3} + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = \frac{π}{3} + 2 π 1

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

改良

1.73205 \dots = 2.11695 \dots

\Rightarrow 偽

解答を確認する

\frac{5 π}{3} + 2 πn :

偽

\frac{5 π}{3} + 2 πn

挿入

n = 1

\frac{5 π}{3} + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = \frac{5 π}{3} + 2 π 1

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

改良

- 1.73205 \dots = - 2.11695 \dots

\Rightarrow 偽

解答を確認する

\frac{2 π}{3} + 2 πn :

真

\frac{2 π}{3} + 2 πn

挿入

n = 1

\frac{2 π}{3} + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = \frac{2 π}{3} + 2 π 1

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

改良

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

解答を確認する

\frac{4 π}{3} + 2 πn :

真

\frac{4 π}{3} + 2 πn

挿入

n = 1

\frac{4 π}{3} + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = \frac{4 π}{3} + 2 π 1

tan (\frac{4 π}{3} + 2 π 1) = \frac{5 + cos ( \frac{4 π}{3} + 2 π 1 )}{6 sin ( \frac{4 π}{3} + 2 π 1 ) cos ( \frac{4 π}{3} + 2 π 1 )}

改良

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

解答を確認する

arccos (\frac{1}{3}) + 2 πn :

真

arccos (\frac{1}{3}) + 2 πn

挿入

n = 1

arccos (\frac{1}{3}) + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = arccos (\frac{1}{3}) + 2 π 1

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

改良

2.82842 \dots = 2.82842 \dots

\Rightarrow 真

解答を確認する

2 π - arccos (\frac{1}{3}) + 2 πn :

真

2 π - arccos (\frac{1}{3}) + 2 πn

挿入

n = 1

2 π - arccos (\frac{1}{3}) + 2 π 1

tan (x) = \frac{5 + cos ( x )}{6 sin ( x ) cos ( x )}

の挿入向け

x = 2 π - arccos (\frac{1}{3}) + 2 π 1

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

改良

- 2.82842 \dots = - 2.82842 \dots

\Rightarrow 真

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

10進法形式で解を証明する

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

グラフ

人気の例

cos (t) = \frac{21}{29}

cos(x)=sin(x-pi/3)

cos (x) = sin (x - \frac{π}{3})

sin(x+pi/4)+sin(x+pi/4)=-1

sin (x + \frac{π}{4}) + sin (x + \frac{π}{4}) = - 1

4 sin^{2} (x) + 9 = 12

tan(-60s)=-tan(60)

tan (- 60 s) = - tan (6 0^{\circ})