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人気のある三角関数 >

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

解

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

解

x = 0.67351 \dots + πn

+1

度

x = 38.58935 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

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

両辺から

3 tan (x)

を引く

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

三角関数の公式を使用して書き換える

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

2倍角の公式を使用:

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

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

分数を組み合わせる

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

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

元を分数に変換する:

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

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

拡張

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

マイナス・プラスの規則を適用する

- (- a) = a

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

簡素化

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

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

3 \cdot 1 \cdot tan (x) = 3 tan (x)

3 \cdot 1 \cdot tan (x)

数を乗じる:

3 \cdot 1 = 3

= 3 tan (x)

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

3 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)

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

数を足す:

2 + 1 = 3

= 3 tan^{3} (x)

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

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

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

類似した元を足す:

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

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

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

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

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

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

置換で解く

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

仮定:

tan (x) = u

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

- 2 + \frac{- u + 3 u ^{3}}{1 - u ^{2}} = 0 : u \approx 0.79798 \dots

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

以下で両辺を乗じる:

1 - u^{2}

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

以下で両辺を乗じる:

1 - u^{2}

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

簡素化

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

簡素化

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

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

分数を乗じる:

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

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

共通因数を約分する:

1 - u^{2}

= - - u + 3 u^{3}

簡素化

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

0 \cdot (1 - u^{2})

規則を適用

0 \cdot a = 0

= 0

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

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

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

解く

- 2 (1 - u^{2}) - u + 3 u^{3} = 0 : u \approx 0.79798 \dots

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

拡張

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

- 2 (1 - u^{2}) - u + 3 u^{3}

拡張

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

- 2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

マイナス・プラスの規則を適用する

- (- a) = a

= - 2 \cdot 1 + 2 u^{2}

数を乗じる:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

ニュートン・ラプソン法を使用して

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

の解を1つ求める:

u \approx 0.79798 \dots

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

ニュートン・ラプソン概算の定義

f (u) = 3 u^{3} + 2 u^{2} - u - 2

発見する

f^{^{'}} (u) : 9 u^{2} + 4 u - 1

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (3 u^{3}) + \frac{d}{d u} (2 u^{2}) - \frac{d u}{d u} - \frac{d}{d u} (2)

\frac{d}{d u} (3 u^{3}) = 9 u^{2}

\frac{d}{d u} (3 u^{3})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{3})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 \cdot 3 u^{3 - 1}

簡素化

= 9 u^{2}

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 u^{2 - 1}

簡素化

= 4 u

\frac{d u}{d u} = 1

\frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 1

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= 9 u^{2} + 4 u - 1 - 0

簡素化

= 9 u^{2} + 4 u - 1

仮定:

u_{0} = 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = 0.83333 \dots : Δ u_{1} = 0.16666 \dots

f (u_{0}) = 3 \cdot 1^{3} + 2 \cdot 1^{2} - 1 - 2 = 2

f^{^{'}} (u_{0}) = 9 \cdot 1^{2} + 4 \cdot 1 - 1 = 12

u_{1} = 0.83333 \dots

Δ u_{1} = ∣ 0.83333 \dots - 1 ∣ = 0.16666 \dots

Δ u_{1} = 0.16666 \dots

u_{2} = 0.79935 \dots : Δ u_{2} = 0.03398 \dots

f (u_{1}) = 3 \cdot 0.83333 \dots^{3} + 2 \cdot 0.83333 \dots^{2} - 0.83333 \dots - 2 = 0.29166 \dots

f^{^{'}} (u_{1}) = 9 \cdot 0.83333 \dots^{2} + 4 \cdot 0.83333 \dots - 1 = 8.58333 \dots

u_{2} = 0.79935 \dots

Δ u_{2} = ∣ 0.79935 \dots - 0.83333 \dots ∣ = 0.03398 \dots

Δ u_{2} = 0.03398 \dots

u_{3} = 0.79798 \dots : Δ u_{3} = 0.00136 \dots

f (u_{2}) = 3 \cdot 0.79935 \dots^{3} + 2 \cdot 0.79935 \dots^{2} - 0.79935 \dots - 2 = 0.01085 \dots

f^{^{'}} (u_{2}) = 9 \cdot 0.79935 \dots^{2} + 4 \cdot 0.79935 \dots - 1 = 7.94809 \dots

u_{3} = 0.79798 \dots

Δ u_{3} = ∣ 0.79798 \dots - 0.79935 \dots ∣ = 0.00136 \dots

Δ u_{3} = 0.00136 \dots

u_{4} = 0.79798 \dots : Δ u_{4} = 2.16223 E - 6

f (u_{3}) = 3 \cdot 0.79798 \dots^{3} + 2 \cdot 0.79798 \dots^{2} - 0.79798 \dots - 2 = 0.00001 \dots

f^{^{'}} (u_{3}) = 9 \cdot 0.79798 \dots^{2} + 4 \cdot 0.79798 \dots - 1 = 7.92300 \dots

u_{4} = 0.79798 \dots

Δ u_{4} = ∣ 0.79798 \dots - 0.79798 \dots ∣ = 2.16223 E - 6

Δ u_{4} = 2.16223 E - 6

u_{5} = 0.79798 \dots : Δ u_{5} = 5.41816 E - 12

f (u_{4}) = 3 \cdot 0.79798 \dots^{3} + 2 \cdot 0.79798 \dots^{2} - 0.79798 \dots - 2 = 4.29279 E - 11

f^{^{'}} (u_{4}) = 9 \cdot 0.79798 \dots^{2} + 4 \cdot 0.79798 \dots - 1 = 7.92296 \dots

u_{5} = 0.79798 \dots

Δ u_{5} = ∣ 0.79798 \dots - 0.79798 \dots ∣ = 5.41816 E - 12

Δ u_{5} = 5.41816 E - 12

u \approx 0.79798 \dots

長除法を適用する:

\frac{3 u ^{3} + 2 u ^{2} - u - 2}{u - 0.79798 \dots} = 3 u^{2} + 4.39395 \dots u + 2.50631 \dots

3 u^{2} + 4.39395 \dots u + 2.50631 \dots \approx 0

ニュートン・ラプソン法を使用して

3 u^{2} + 4.39395 \dots u + 2.50631 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

3 u^{2} + 4.39395 \dots u + 2.50631 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = 3 u^{2} + 4.39395 \dots u + 2.50631 \dots

発見する

f^{^{'}} (u) : 6 u + 4.39395 \dots

\frac{d}{d u} (3 u^{2} + 4.39395 \dots u + 2.50631 \dots)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (3 u^{2}) + \frac{d}{d u} (4.39395 \dots u) + \frac{d}{d u} (2.50631 \dots)

\frac{d}{d u} (3 u^{2}) = 6 u

\frac{d}{d u} (3 u^{2})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

簡素化

= 6 u

\frac{d}{d u} (4.39395 \dots u) = 4.39395 \dots

\frac{d}{d u} (4.39395 \dots u)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 4.39395 \dots \frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 4.39395 \dots \cdot 1

簡素化

= 4.39395 \dots

\frac{d}{d u} (2.50631 \dots) = 0

\frac{d}{d u} (2.50631 \dots)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= 6 u + 4.39395 \dots + 0

簡素化

= 6 u + 4.39395 \dots

仮定:

u_{0} = - 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = - 0.30739 \dots : Δ u_{1} = 0.69260 \dots

f (u_{0}) = 3 (- 1)^{2} + 4.39395 \dots (- 1) + 2.50631 \dots = 1.11235 \dots

f^{^{'}} (u_{0}) = 6 (- 1) + 4.39395 \dots = - 1.60604 \dots

u_{1} = - 0.30739 \dots

Δ u_{1} = ∣ - 0.30739 \dots - (- 1) ∣ = 0.69260 \dots

Δ u_{1} = 0.69260 \dots

u_{2} = - 0.87184 \dots : Δ u_{2} = 0.56444 \dots

f (u_{1}) = 3 (- 0.30739 \dots)^{2} + 4.39395 \dots (- 0.30739 \dots) + 2.50631 \dots = 1.43911 \dots

f^{^{'}} (u_{1}) = 6 (- 0.30739 \dots) + 4.39395 \dots = 2.54959 \dots

u_{2} = - 0.87184 \dots

Δ u_{2} = ∣ - 0.87184 \dots - (- 0.30739 \dots) ∣ = 0.56444 \dots

Δ u_{2} = 0.56444 \dots

u_{3} = 0.26997 \dots : Δ u_{3} = 1.14181 \dots

f (u_{2}) = 3 (- 0.87184 \dots)^{2} + 4.39395 \dots (- 0.87184 \dots) + 2.50631 \dots = 0.95580 \dots

f^{^{'}} (u_{2}) = 6 (- 0.87184 \dots) + 4.39395 \dots = - 0.83708 \dots

u_{3} = 0.26997 \dots

Δ u_{3} = ∣ 0.26997 \dots - (- 0.87184 \dots) ∣ = 1.14181 \dots

Δ u_{3} = 1.14181 \dots

u_{4} = - 0.38040 \dots : Δ u_{4} = 0.65037 \dots

f (u_{3}) = 3 \cdot 0.26997 \dots^{2} + 4.39395 \dots \cdot 0.26997 \dots + 2.50631 \dots = 3.91122 \dots

f^{^{'}} (u_{3}) = 6 \cdot 0.26997 \dots + 4.39395 \dots = 6.01380 \dots

u_{4} = - 0.38040 \dots

Δ u_{4} = ∣ - 0.38040 \dots - 0.26997 \dots ∣ = 0.65037 \dots

Δ u_{4} = 0.65037 \dots

u_{5} = - 0.98136 \dots : Δ u_{5} = 0.60096 \dots

f (u_{4}) = 3 (- 0.38040 \dots)^{2} + 4.39395 \dots (- 0.38040 \dots) + 2.50631 \dots = 1.26896 \dots

f^{^{'}} (u_{4}) = 6 (- 0.38040 \dots) + 4.39395 \dots = 2.11155 \dots

u_{5} = - 0.98136 \dots

Δ u_{5} = ∣ - 0.98136 \dots - (- 0.38040 \dots) ∣ = 0.60096 \dots

Δ u_{5} = 0.60096 \dots

u_{6} = - 0.25625 \dots : Δ u_{6} = 0.72510 \dots

f (u_{5}) = 3 (- 0.98136 \dots)^{2} + 4.39395 \dots (- 0.98136 \dots) + 2.50631 \dots = 1.08346 \dots

f^{^{'}} (u_{5}) = 6 (- 0.98136 \dots) + 4.39395 \dots = - 1.49421 \dots

u_{6} = - 0.25625 \dots

Δ u_{6} = ∣ - 0.25625 \dots - (- 0.98136 \dots) ∣ = 0.72510 \dots

Δ u_{6} = 0.72510 \dots

u_{7} = - 0.80846 \dots : Δ u_{7} = 0.55220 \dots

f (u_{6}) = 3 (- 0.25625 \dots)^{2} + 4.39395 \dots (- 0.25625 \dots) + 2.50631 \dots = 1.57733 \dots

f^{^{'}} (u_{6}) = 6 (- 0.25625 \dots) + 4.39395 \dots = 2.85642 \dots

u_{7} = - 0.80846 \dots

Δ u_{7} = ∣ - 0.80846 \dots - (- 0.25625 \dots) ∣ = 0.55220 \dots

Δ u_{7} = 0.55220 \dots

u_{8} = 1.19406 \dots : Δ u_{8} = 2.00252 \dots

f (u_{7}) = 3 (- 0.80846 \dots)^{2} + 4.39395 \dots (- 0.80846 \dots) + 2.50631 \dots = 0.91479 \dots

f^{^{'}} (u_{7}) = 6 (- 0.80846 \dots) + 4.39395 \dots = - 0.45682 \dots

u_{8} = 1.19406 \dots

Δ u_{8} = ∣ 1.19406 \dots - (- 0.80846 \dots) ∣ = 2.00252 \dots

Δ u_{8} = 2.00252 \dots

u_{9} = 0.15322 \dots : Δ u_{9} = 1.04083 \dots

f (u_{8}) = 3 \cdot 1.19406 \dots^{2} + 4.39395 \dots \cdot 1.19406 \dots + 2.50631 \dots = 12.03031 \dots

f^{^{'}} (u_{8}) = 6 \cdot 1.19406 \dots + 4.39395 \dots = 11.55832 \dots

u_{9} = 0.15322 \dots

Δ u_{9} = ∣ 0.15322 \dots - 1.19406 \dots ∣ = 1.04083 \dots

Δ u_{9} = 1.04083 \dots

u_{10} = - 0.45844 \dots : Δ u_{10} = 0.61167 \dots

f (u_{9}) = 3 \cdot 0.15322 \dots^{2} + 4.39395 \dots \cdot 0.15322 \dots + 2.50631 \dots = 3.25001 \dots

f^{^{'}} (u_{9}) = 6 \cdot 0.15322 \dots + 4.39395 \dots = 5.31331 \dots

u_{10} = - 0.45844 \dots

Δ u_{10} = ∣ - 0.45844 \dots - 0.15322 \dots ∣ = 0.61167 \dots

Δ u_{10} = 0.61167 \dots

u_{11} = - 1.14149 \dots : Δ u_{11} = 0.68305 \dots

f (u_{10}) = 3 (- 0.45844 \dots)^{2} + 4.39395 \dots (- 0.45844 \dots) + 2.50631 \dots = 1.12243 \dots

f^{^{'}} (u_{10}) = 6 (- 0.45844 \dots) + 4.39395 \dots = 1.64326 \dots

u_{11} = - 1.14149 \dots

Δ u_{11} = ∣ - 1.14149 \dots - (- 0.45844 \dots) ∣ = 0.68305 \dots

Δ u_{11} = 0.68305 \dots

解を見つけられない

解は

u \approx 0.79798 \dots

u \approx 0.79798 \dots

解を検算する

未定義の (特異) 点を求める:

u = 1, u = - 1

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

の分母をゼロに比較する

解く

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

1 - u^{2} = 0

1

を右側に移動します

1 - u^{2} = 0

両辺から

1

を引く

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

簡素化

- u^{2} = - 1

- u^{2} = - 1

以下で両辺を割る

- 1

- u^{2} = - 1

以下で両辺を割る

- 1

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

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

累乗根の規則を適用する:

1 = 1

= 1

- 1 = - 1

- 1

累乗根の規則を適用する:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下の点は定義されていない

u = 1, u = - 1

未定義のポイントを解に組み合わせる:

u \approx 0.79798 \dots

代用を戻す

u = tan (x)

tan (x) \approx 0.79798 \dots

tan (x) \approx 0.79798 \dots

tan (x) = 0.79798 \dots : x = arctan (0.79798 \dots) + πn

tan (x) = 0.79798 \dots

三角関数の逆数プロパティを適用する

tan (x) = 0.79798 \dots

以下の一般解

tan (x) = 0.79798 \dots

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

x = arctan (0.79798 \dots) + πn

x = arctan (0.79798 \dots) + πn

すべての解を組み合わせる

x = arctan (0.79798 \dots) + πn

10進法形式で解を証明する

x = 0.67351 \dots + πn

グラフ

人気の例

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arccot (x - 2) = arccot (x - 1) + arccot (x)

arccos (2 x) = π

solvefor t,x=4sin(t)

so l v e f or t, x = 4 sin (t)

sin(2x)+1.5cos(x)=0

sin (2 x) + 1.5 cos (x) = 0