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人気のある三角関数 >

3tan(2x)-3cot(x)=0

解

3 tan (2 x) - 3 cot (x) = 0

解

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

+1

度

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

解答ステップ

3 tan (2 x) - 3 cot (x) = 0

三角関数の公式を使用して書き換える

- 3 cot (x) + 3 tan (2 x)

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

tan (2 x)

2倍角の公式を使用:

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

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

因数

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

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

因数

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

1 - tan^{2} (x)

2乗の差の公式を適用する:

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

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

= (1 + tan (x)) (1 - tan (x))

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

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

= - 3 cot (x) + 3 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

3 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} = \frac{6 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

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

分数を乗じる:

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

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

数を乗じる:

2 \cdot 3 = 6

= \frac{6 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )}

= - 3 cot (x) + \frac{6 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

基本的な三角関数の公式を使用する:

cot (x) = \frac{1}{tan ( x )}

= \frac{6 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 3 \cdot \frac{1}{tan ( x )}

簡素化

\frac{6 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 3 \cdot \frac{1}{tan ( x )} : - \frac{9 tan ^{2} ( x ) - 3}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

\frac{6 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 3 \cdot \frac{1}{tan ( x )}

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

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

分数を乗じる:

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

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

数を乗じる:

1 \cdot 3 = 3

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

= \frac{6 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )} - \frac{3}{tan ( x )}

因数

(1 + tan (x)) (1 - tan (x)) : - (1 + tan (x)) (tan (x) - 1)

(1 + tan (x)) (1 - tan (x))

因数

1 - tan (x) : - (tan (x) - 1)

1 - tan (x)

共通項をくくり出す

- 1

= - (tan (x) - 1)

= - (1 + tan (x)) (tan (x) - 1)

= \frac{6 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} - \frac{3}{tan ( x )}

以下の最小公倍数:

- (1 + tan (x)) (tan (x) - 1), tan (x) : - tan (x) (tan (x) + 1) (tan (x) - 1)

- (1 + tan (x)) (tan (x) - 1), tan (x)

最小公倍数 (LCM)

- (1 + tan (x)) (tan (x) - 1)

または以下のいずれかに現れる因数で構成された式を計算する:

tan (x)

= - tan (x) (tan (x) + 1) (tan (x) - 1)

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

- tan (x) (tan (x) + 1) (tan (x) - 1)

\frac{6 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )}

の場合

:

分母と分子に以下を乗じる:

tan (x)

\frac{6 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} = \frac{6 tan ( x ) tan ( x )}{( - ( 1 + tan ( x ) ) ( tan ( x ) - 1 ) ) tan ( x )} = \frac{6 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

\frac{3}{tan ( x )}

の場合

:

分母と分子に以下を乗じる:

- (tan (x) + 1) (tan (x) - 1)

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

= \frac{6 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )} - \frac{- 3 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

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

改良

= - \frac{6 tan ^{2} ( x ) + 3 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

拡張

6 tan^{2} (x) + 3 (tan (x) + 1) (tan (x) - 1) : 9 tan^{2} (x) - 3

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

拡張

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

拡張

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

(tan (x) + 1) (tan (x) - 1)

2乗の差の公式を適用する:

(a + b) (a - b) = a^{2} - b^{2}

a = tan (x), b = 1

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

規則を適用

1^{a} = 1

1^{2} = 1

= tan^{2} (x) - 1

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

3 \cdot 1 = 3

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

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

= 6 tan^{2} (x) + 3 tan^{2} (x) - 3

類似した元を足す:

6 tan^{2} (x) + 3 tan^{2} (x) = 9 tan^{2} (x)

= 9 tan^{2} (x) - 3

= - \frac{9 tan ^{2} ( x ) - 3}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

= - \frac{9 tan ^{2} ( x ) - 3}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

- \frac{- 3 + 9 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

置換で解く

- \frac{- 3 + 9 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

仮定:

tan (x) = u

- \frac{- 3 + 9 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

- \frac{- 3 + 9 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- \frac{- 3 + 9 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

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

- 3 + 9 u^{2} = 0

解く

- 3 + 9 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- 3 + 9 u^{2} = 0

3

を右側に移動します

- 3 + 9 u^{2} = 0

両辺に

3

を足す

- 3 + 9 u^{2} + 3 = 0 + 3

簡素化

9 u^{2} = 3

9 u^{2} = 3

以下で両辺を割る

9

9 u^{2} = 3

以下で両辺を割る

9

\frac{9 u ^{2}}{9} = \frac{3}{9}

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}

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

解を検算する

未定義の (特異) 点を求める:

u = 1, u = - 1, u = 0

- \frac{- 3 + 9 u ^{2}}{( - 1 + u ) ( 1 + u ) u}

の分母をゼロに比較する

解く

(- 1 + u) (1 + u) u = 0 : u = 1, u = - 1, u = 0

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

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

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

解く

- 1 + u = 0 : u = 1

- 1 + u = 0

1

を右側に移動します

- 1 + u = 0

両辺に

1

を足す

- 1 + u + 1 = 0 + 1

簡素化

u = 1

u = 1

解く

1 + u = 0 : u = - 1

1 + u = 0

1

を右側に移動します

1 + u = 0

両辺から

1

を引く

1 + u - 1 = 0 - 1

簡素化

u = - 1

u = - 1

解答は

u = 1, u = - 1, u = 0

以下の点は定義されていない

u = 1, u = - 1, u = 0

未定義のポイントを解に組み合わせる:

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

代用を戻す

u = tan (x)

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

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

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

グラフ

人気の例

cos (θ) = (\frac{1}{4})

(36)/(sin(110))=(15)/(sin(x))

\frac{36}{sin ( 11 0 ^{\circ} )} = \frac{15}{sin ( x )}

6sin(x/2)+6cos(x)=0

6 sin (\frac{x}{2}) + 6 cos (x) = 0

cos (x) = 0.925

(sin(A))/9 =(sin(108))/6

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