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人気のある三角関数 >

solvefor x,tan(3x)=5tan(x)

解

解く

x, tan (3 x) = 5 tan (x)

解

x = πn, x = - 0.36136 \dots + πn, x = 0.36136 \dots + πn

+1

度

x = 0^{\circ} + 18 0^{\circ} n, x = - 20.70481 \dots^{\circ} + 18 0^{\circ} n, x = 20.70481 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

tan (3 x) = 5 tan (x)

両辺から

5 tan (x)

を引く

tan (3 x) - 5 tan (x) = 0

三角関数の公式を使用して書き換える

tan (3 x) - 5 tan (x)

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

tan (3 x)

三角関数の公式を使用して書き換える

tan (3 x)

書き換え

= tan (2 x + x)

角の和の公式を使用する:

tan (s + t) = \frac{tan ( s ) + tan ( t )}{1 - tan ( s ) tan ( t )}

= \frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )}

= \frac{tan ( 2 x ) + tan ( x )}{1 - tan ( 2 x ) tan ( x )}

2倍角の公式を使用:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= \frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

簡素化

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )} : \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} tan (x) = \frac{2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} tan (x)

分数を乗じる:

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

= \frac{2 tan ( x ) tan ( x )}{1 - tan ^{2} ( x )}

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

2 tan (x) tan (x)

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

= 2 tan^{2} (x)

= \frac{2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

= \frac{\frac{2 t a n ( x )}{- t a n ^{2} ( x ) + 1} + tan ( x )}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

結合

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x) : \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - tan ^{2} ( x )}

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} + tan (x)

元を分数に変換する:

tan (x) = \frac{tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

= \frac{2 tan ( x )}{1 - tan ^{2} ( x )} + \frac{tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{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{2 tan ( x ) + tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

拡張

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

2 tan (x) + tan (x) (1 - tan^{2} (x))

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

= tan (x) 1 - tan (x) tan^{2} (x)

= 1 tan (x) - tan^{2} (x) tan (x)

簡素化

1 \cdot tan (x) - tan^{2} (x) tan (x) : tan (x) - tan^{3} (x)

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

1 \cdot tan (x) = tan (x)

1 tan (x)

乗算:

1 \cdot tan (x) = tan (x)

= tan (x)

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

tan^{2} (x) tan (x)

指数の規則を適用する:

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

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

= tan^{2 + 1} (x)

数を足す:

2 + 1 = 3

= tan^{3} (x)

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

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

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

類似した元を足す:

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

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

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

= \frac{\frac{3 t a n ( x ) - t a n ^{3} ( x )}{1 - t a n ^{2} ( x )}}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

分数の規則を適用する:

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{( 1 - tan ^{2} ( x ) ) ( 1 - \frac{2 t a n ^{2} ( x )}{1 - t a n ^{2} ( x )} )}

結合

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

1 - \frac{2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

元を分数に変換する:

1 = \frac{1 ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

= \frac{1 ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )} - \frac{2 tan ^{2} ( x )}{1 - tan ^{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{1 ( 1 - tan ^{2} ( x ) ) - 2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

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

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

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

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

乗算:

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

= 1 - tan^{2} (x)

括弧を削除する:

(a) = a

= 1 - tan^{2} (x)

= 1 - tan^{2} (x) - 2 tan^{2} (x)

類似した元を足す:

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

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

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{\frac{- 3 t a n ^{2} ( x ) + 1}{- t a n ^{2} ( x ) + 1} ( - tan ^{2} ( x ) + 1 )}

乗じる

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

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

分数を乗じる:

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

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

共通因数を約分する:

1 - tan^{2} (x)

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

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

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x)

簡素化

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x) : \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x)

元を分数に変換する:

5 tan (x) = \frac{5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - \frac{5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{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{3 tan ( x ) - tan ^{3} ( x ) - 5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

拡張

3 tan (x) - tan^{3} (x) - 5 tan (x) (1 - 3 tan^{2} (x)) : - 2 tan (x) + 14 tan^{3} (x)

3 tan (x) - tan^{3} (x) - 5 tan (x) (1 - 3 tan^{2} (x))

拡張

- 5 tan (x) (1 - 3 tan^{2} (x)) : - 5 tan (x) + 15 tan^{3} (x)

- 5 tan (x) (1 - 3 tan^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 5 tan (x), b = 1, c = 3 tan^{2} (x)

= - 5 tan (x) \cdot 1 - (- 5 tan (x)) \cdot 3 tan^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x)

簡素化

- 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x) : - 5 tan (x) + 15 tan^{3} (x)

- 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x)

5 \cdot 1 \cdot tan (x) = 5 tan (x)

5 \cdot 1 \cdot tan (x)

数を乗じる:

5 \cdot 1 = 5

= 5 tan (x)

5 \cdot 3 tan^{2} (x) tan (x) = 15 tan^{3} (x)

5 \cdot 3 tan^{2} (x) tan (x)

数を乗じる:

5 \cdot 3 = 15

= 15 tan^{2} (x) tan (x)

指数の規則を適用する:

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

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

= 15 tan^{2 + 1} (x)

数を足す:

2 + 1 = 3

= 15 tan^{3} (x)

= - 5 tan (x) + 15 tan^{3} (x)

= - 5 tan (x) + 15 tan^{3} (x)

= 3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x)

簡素化

3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x) : - 2 tan (x) + 14 tan^{3} (x)

3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x)

類似した元を足す:

- tan^{3} (x) + 15 tan^{3} (x) = 14 tan^{3} (x)

= 3 tan (x) + 14 tan^{3} (x) - 5 tan (x)

類似した元を足す:

3 tan (x) - 5 tan (x) = - 2 tan (x)

= - 2 tan (x) + 14 tan^{3} (x)

= - 2 tan (x) + 14 tan^{3} (x)

= \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

= \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{14 tan ^{3} ( x ) - 2 tan ( x )}{1 - 3 tan ^{2} ( x )} = 0

置換で解く

\frac{14 tan ^{3} ( x ) - 2 tan ( x )}{1 - 3 tan ^{2} ( x )} = 0

仮定:

tan (x) = u

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0 : u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0

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

14 u^{3} - 2 u = 0

解く

14 u^{3} - 2 u = 0 : u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

14 u^{3} - 2 u = 0

因数

14 u^{3} - 2 u : 2 u (7 u + 1) (7 u - 1)

14 u^{3} - 2 u

共通項をくくり出す

2 u : 2 u (7 u^{2} - 1)

14 u^{3} - 2 u

指数の規則を適用する:

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

u^{3} = u^{2} u

= 14 u^{2} u - 2 u

14

を書き換え

2 \cdot 7

= 2 \cdot 7 u^{2} u - 2 u

共通項をくくり出す

2 u

= 2 u (7 u^{2} - 1)

= 2 u (7 u^{2} - 1)

因数

7 u^{2} - 1 : (7 u + 1) (7 u - 1)

7 u^{2} - 1

7 u^{2} - 1

を書き換え

(7 u)^{2} - 1^{2}

7 u^{2} - 1

累乗根の規則を適用する:

a = (a)^{2}

7 = (7)^{2}

= (7)^{2} u^{2} - 1

1

を書き換え

1^{2}

= (7)^{2} u^{2} - 1^{2}

指数の規則を適用する:

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

(7)^{2} u^{2} = (7 u)^{2}

= (7 u)^{2} - 1^{2}

= (7 u)^{2} - 1^{2}

2乗の差の公式を適用する:

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

(7 u)^{2} - 1^{2} = (7 u + 1) (7 u - 1)

= (7 u + 1) (7 u - 1)

= 2 u (7 u + 1) (7 u - 1)

2 u (7 u + 1) (7 u - 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

u = 0 or 7 u + 1 = 0 or 7 u - 1 = 0

解く

7 u + 1 = 0 : u = - \frac{7}{7}

7 u + 1 = 0

1

を右側に移動します

7 u + 1 = 0

両辺から

1

を引く

7 u + 1 - 1 = 0 - 1

簡素化

7 u = - 1

7 u = - 1

以下で両辺を割る

7

7 u = - 1

以下で両辺を割る

7

\frac{7 u}{7} = \frac{- 1}{7}

簡素化

\frac{7 u}{7} = \frac{- 1}{7}

簡素化

\frac{7 u}{7} : u

\frac{7 u}{7}

共通因数を約分する:

7

= u

簡素化

\frac{- 1}{7} : - \frac{7}{7}

\frac{- 1}{7}

分数の規則を適用する:

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

= - \frac{1}{7}

有理化する

- \frac{1}{7} : - \frac{7}{7}

- \frac{1}{7}

共役で乗じる

\frac{7}{7}

= - \frac{1 \cdot 7}{7 7}

1 \cdot 7 = 7

77 = 7

77

累乗根の規則を適用する:

a a = a

77 = 7

= 7

= - \frac{7}{7}

= - \frac{7}{7}

u = - \frac{7}{7}

u = - \frac{7}{7}

u = - \frac{7}{7}

解く

7 u - 1 = 0 : u = \frac{7}{7}

7 u - 1 = 0

1

を右側に移動します

7 u - 1 = 0

両辺に

1

を足す

7 u - 1 + 1 = 0 + 1

簡素化

7 u = 1

7 u = 1

以下で両辺を割る

7

7 u = 1

以下で両辺を割る

7

\frac{7 u}{7} = \frac{1}{7}

簡素化

\frac{7 u}{7} = \frac{1}{7}

簡素化

\frac{7 u}{7} : u

\frac{7 u}{7}

共通因数を約分する:

7

= u

簡素化

\frac{1}{7} : \frac{7}{7}

\frac{1}{7}

共役で乗じる

\frac{7}{7}

= \frac{1 \cdot 7}{7 7}

1 \cdot 7 = 7

77 = 7

77

累乗根の規則を適用する:

a a = a

77 = 7

= 7

= \frac{7}{7}

u = \frac{7}{7}

u = \frac{7}{7}

u = \frac{7}{7}

解答は

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

解を検算する

未定義の (特異) 点を求める:

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

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}}

の分母をゼロに比較する

解く

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

1 - 3 u^{2} = 0

1

を右側に移動します

1 - 3 u^{2} = 0

両辺から

1

を引く

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

簡素化

- 3 u^{2} = - 1

- 3 u^{2} = - 1

以下で両辺を割る

- 3

- 3 u^{2} = - 1

以下で両辺を割る

- 3

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

簡素化

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

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

\frac{1}{3} = \frac{1}{3}

\frac{1}{3}

累乗根の規則を適用する:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

累乗根の規則を適用する:

1 = 1

1 = 1

= \frac{1}{3}

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

- \frac{1}{3}

累乗根の規則を適用する:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

累乗根の規則を適用する:

1 = 1

1 = 1

= - \frac{1}{3}

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

以下の点は定義されていない

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

未定義のポイントを解に組み合わせる:

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

代用を戻す

u = tan (x)

tan (x) = 0, tan (x) = - \frac{7}{7}, tan (x) = \frac{7}{7}

tan (x) = 0, tan (x) = - \frac{7}{7}, tan (x) = \frac{7}{7}

tan (x) = 0 : x = πn

tan (x) = 0

以下の一般解

tan (x) = 0

tan (x)

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

解く

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - \frac{7}{7} : x = arctan (- \frac{7}{7}) + πn

tan (x) = - \frac{7}{7}

三角関数の逆数プロパティを適用する

tan (x) = - \frac{7}{7}

以下の一般解

tan (x) = - \frac{7}{7}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{7}{7}) + πn

x = arctan (- \frac{7}{7}) + πn

tan (x) = \frac{7}{7} : x = arctan (\frac{7}{7}) + πn

tan (x) = \frac{7}{7}

三角関数の逆数プロパティを適用する

tan (x) = \frac{7}{7}

以下の一般解

tan (x) = \frac{7}{7}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{7}{7}) + πn

x = arctan (\frac{7}{7}) + πn

すべての解を組み合わせる

x = πn, x = arctan (- \frac{7}{7}) + πn, x = arctan (\frac{7}{7}) + πn

10進法形式で解を証明する

x = πn, x = - 0.36136 \dots + πn, x = 0.36136 \dots + πn

グラフ

人気の例

3 sin (2 t) = 0

tan (θ) = \frac{8}{5}

tan (θ) = \frac{8}{8}

cos(2x-pi/(14))=0

cos (2 x - \frac{π}{14}) = 0

2 = 2 tan (x)