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人気のある >

cos^3(x)+sin^2(x)=0

解

cos^{3} (x) + sin^{2} (x) = 0

解

x = 2.42626 \dots + 2 πn, x = - 2.42626 \dots + 2 πn

+1

度

x = 139.01468 \dots^{\circ} + 36 0^{\circ} n, x = - 139.01468 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

cos^{3} (x) + sin^{2} (x) = 0

三角関数の公式を使用して書き換える

cos^{3} (x) + 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)

= cos^{3} (x) + 1 - cos^{2} (x)

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

置換で解く

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

仮定:

cos (x) = u

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

1 - u^{2} + u^{3} = 0 : u \approx - 0.75487 \dots

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

標準的な形式で書く

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

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

ニュートン・ラプソン法を使用して

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

の解を1つ求める:

u \approx - 0.75487 \dots

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

ニュートン・ラプソン概算の定義

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

発見する

f^{^{'}} (u) : 3 u^{2} - 2 u

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

和/差の法則を適用:

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

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

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

\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 u^{3 - 1}

簡素化

= 3 u^{2}

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

\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 u^{2 - 1}

簡素化

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

定数の導関数:

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

= 0

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

簡素化

= 3 u^{2} - 2 u

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} = - 0.8 : Δ u_{1} = 0.2

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

f^{^{'}} (u_{0}) = 3 (- 1)^{2} - 2 (- 1) = 5

u_{1} = - 0.8

Δ u_{1} = ∣ - 0.8 - (- 1) ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = - 0.75681 \dots : Δ u_{2} = 0.04318 \dots

f (u_{1}) = (- 0.8)^{3} - (- 0.8)^{2} + 1 = - 0.152

f^{^{'}} (u_{1}) = 3 (- 0.8)^{2} - 2 (- 0.8) = 3.52

u_{2} = - 0.75681 \dots

Δ u_{2} = ∣ - 0.75681 \dots - (- 0.8) ∣ = 0.04318 \dots

Δ u_{2} = 0.04318 \dots

u_{3} = - 0.75488 \dots : Δ u_{3} = 0.00193 \dots

f (u_{2}) = (- 0.75681 \dots)^{3} - (- 0.75681 \dots)^{2} + 1 = - 0.00625 \dots

f^{^{'}} (u_{2}) = 3 (- 0.75681 \dots)^{2} - 2 (- 0.75681 \dots) = 3.23195 \dots

u_{3} = - 0.75488 \dots

Δ u_{3} = ∣ - 0.75488 \dots - (- 0.75681 \dots) ∣ = 0.00193 \dots

Δ u_{3} = 0.00193 \dots

u_{4} = - 0.75487 \dots : Δ u_{4} = 3.80818 E - 6

f (u_{3}) = (- 0.75488 \dots)^{3} - (- 0.75488 \dots)^{2} + 1 = - 0.00001 \dots

f^{^{'}} (u_{3}) = 3 (- 0.75488 \dots)^{2} - 2 (- 0.75488 \dots) = 3.21930 \dots

u_{4} = - 0.75487 \dots

Δ u_{4} = ∣ - 0.75487 \dots - (- 0.75488 \dots) ∣ = 3.80818 E - 6

Δ u_{4} = 3.80818 E - 6

u_{5} = - 0.75487 \dots : Δ u_{5} = 1.47065 E - 11

f (u_{4}) = (- 0.75487 \dots)^{3} - (- 0.75487 \dots)^{2} + 1 = - 4.73444 E - 11

f^{^{'}} (u_{4}) = 3 (- 0.75487 \dots)^{2} - 2 (- 0.75487 \dots) = 3.21927 \dots

u_{5} = - 0.75487 \dots

Δ u_{5} = ∣ - 0.75487 \dots - (- 0.75487 \dots) ∣ = 1.47065 E - 11

Δ u_{5} = 1.47065 E - 11

u \approx - 0.75487 \dots

長除法を適用する:

\frac{u ^{3} - u ^{2} + 1}{u + 0.75487 \dots} = u^{2} - 1.75487 \dots u + 1.32471 \dots

u^{2} - 1.75487 \dots u + 1.32471 \dots \approx 0

ニュートン・ラプソン法を使用して

u^{2} - 1.75487 \dots u + 1.32471 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

u^{2} - 1.75487 \dots u + 1.32471 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = u^{2} - 1.75487 \dots u + 1.32471 \dots

発見する

f^{^{'}} (u) : 2 u - 1.75487 \dots

\frac{d}{d u} (u^{2} - 1.75487 \dots u + 1.32471 \dots)

和/差の法則を適用:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1.75487 \dots u) + \frac{d}{d u} (1.32471 \dots)

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

\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 u^{2 - 1}

簡素化

= 2 u

\frac{d}{d u} (1.75487 \dots u) = 1.75487 \dots

\frac{d}{d u} (1.75487 \dots u)

定数を除去:

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

= 1.75487 \dots \frac{d u}{d u}

共通の導関数を適用:

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

= 1.75487 \dots \cdot 1

簡素化

= 1.75487 \dots

\frac{d}{d u} (1.32471 \dots) = 0

\frac{d}{d u} (1.32471 \dots)

定数の導関数:

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

= 0

= 2 u - 1.75487 \dots + 0

簡素化

= 2 u - 1.75487 \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} = - 1.32471 \dots : Δ u_{1} = 2.32471 \dots

f (u_{0}) = 1^{2} - 1.75487 \dots \cdot 1 + 1.32471 \dots = 0.56984 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 1.75487 \dots = 0.24512 \dots

u_{1} = - 1.32471 \dots

Δ u_{1} = ∣ - 1.32471 \dots - 1 ∣ = 2.32471 \dots

Δ u_{1} = 2.32471 \dots

u_{2} = - 0.09766 \dots : Δ u_{2} = 1.22705 \dots

f (u_{1}) = (- 1.32471 \dots)^{2} - 1.75487 \dots (- 1.32471 \dots) + 1.32471 \dots = 5.40431 \dots

f^{^{'}} (u_{1}) = 2 (- 1.32471 \dots) - 1.75487 \dots = - 4.40431 \dots

u_{2} = - 0.09766 \dots

Δ u_{2} = ∣ - 0.09766 \dots - (- 1.32471 \dots) ∣ = 1.22705 \dots

Δ u_{2} = 1.22705 \dots

u_{3} = 0.67437 \dots : Δ u_{3} = 0.77204 \dots

f (u_{2}) = (- 0.09766 \dots)^{2} - 1.75487 \dots (- 0.09766 \dots) + 1.32471 \dots = 1.50565 \dots

f^{^{'}} (u_{2}) = 2 (- 0.09766 \dots) - 1.75487 \dots = - 1.95021 \dots

u_{3} = 0.67437 \dots

Δ u_{3} = ∣ 0.67437 \dots - (- 0.09766 \dots) ∣ = 0.77204 \dots

Δ u_{3} = 0.77204 \dots

u_{4} = 2.14204 \dots : Δ u_{4} = 1.46766 \dots

f (u_{3}) = 0.67437 \dots^{2} - 1.75487 \dots \cdot 0.67437 \dots + 1.32471 \dots = 0.59605 \dots

f^{^{'}} (u_{3}) = 2 \cdot 0.67437 \dots - 1.75487 \dots = - 0.40612 \dots

u_{4} = 2.14204 \dots

Δ u_{4} = ∣ 2.14204 \dots - 0.67437 \dots ∣ = 1.46766 \dots

Δ u_{4} = 1.46766 \dots

u_{5} = 1.29037 \dots : Δ u_{5} = 0.85166 \dots

f (u_{4}) = 2.14204 \dots^{2} - 1.75487 \dots \cdot 2.14204 \dots + 1.32471 \dots = 2.15403 \dots

f^{^{'}} (u_{4}) = 2 \cdot 2.14204 \dots - 1.75487 \dots = 2.52920 \dots

u_{5} = 1.29037 \dots

Δ u_{5} = ∣ 1.29037 \dots - 2.14204 \dots ∣ = 0.85166 \dots

Δ u_{5} = 0.85166 \dots

u_{6} = 0.41210 \dots : Δ u_{6} = 0.87826 \dots

f (u_{5}) = 1.29037 \dots^{2} - 1.75487 \dots \cdot 1.29037 \dots + 1.32471 \dots = 0.72533 \dots

f^{^{'}} (u_{5}) = 2 \cdot 1.29037 \dots - 1.75487 \dots = 0.82587 \dots

u_{6} = 0.41210 \dots

Δ u_{6} = ∣ 0.41210 \dots - 1.29037 \dots ∣ = 0.87826 \dots

Δ u_{6} = 0.87826 \dots

u_{7} = 1.24093 \dots : Δ u_{7} = 0.82882 \dots

f (u_{6}) = 0.41210 \dots^{2} - 1.75487 \dots \cdot 0.41210 \dots + 1.32471 \dots = 0.77135 \dots

f^{^{'}} (u_{6}) = 2 \cdot 0.41210 \dots - 1.75487 \dots = - 0.93065 \dots

u_{7} = 1.24093 \dots

Δ u_{7} = ∣ 1.24093 \dots - 0.41210 \dots ∣ = 0.82882 \dots

Δ u_{7} = 0.82882 \dots

u_{8} = 0.29600 \dots : Δ u_{8} = 0.94492 \dots

f (u_{7}) = 1.24093 \dots^{2} - 1.75487 \dots \cdot 1.24093 \dots + 1.32471 \dots = 0.68694 \dots

f^{^{'}} (u_{7}) = 2 \cdot 1.24093 \dots - 1.75487 \dots = 0.72698 \dots

u_{8} = 0.29600 \dots

Δ u_{8} = ∣ 0.29600 \dots - 1.24093 \dots ∣ = 0.94492 \dots

Δ u_{8} = 0.94492 \dots

u_{9} = 1.06383 \dots : Δ u_{9} = 0.76782 \dots

f (u_{8}) = 0.29600 \dots^{2} - 1.75487 \dots \cdot 0.29600 \dots + 1.32471 \dots = 0.89288 \dots

f^{^{'}} (u_{8}) = 2 \cdot 0.29600 \dots - 1.75487 \dots = - 1.16286 \dots

u_{9} = 1.06383 \dots

Δ u_{9} = ∣ 1.06383 \dots - 0.29600 \dots ∣ = 0.76782 \dots

Δ u_{9} = 0.76782 \dots

u_{10} = - 0.51763 \dots : Δ u_{10} = 1.58147 \dots

f (u_{9}) = 1.06383 \dots^{2} - 1.75487 \dots \cdot 1.06383 \dots + 1.32471 \dots = 0.58956 \dots

f^{^{'}} (u_{9}) = 2 \cdot 1.06383 \dots - 1.75487 \dots = 0.37279 \dots

u_{10} = - 0.51763 \dots

Δ u_{10} = ∣ - 0.51763 \dots - 1.06383 \dots ∣ = 1.58147 \dots

Δ u_{10} = 1.58147 \dots

解を見つけられない

解は

u \approx - 0.75487 \dots

代用を戻す

u = cos (x)

cos (x) \approx - 0.75487 \dots

cos (x) \approx - 0.75487 \dots

cos (x) = - 0.75487 \dots : x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

cos (x) = - 0.75487 \dots

三角関数の逆数プロパティを適用する

cos (x) = - 0.75487 \dots

以下の一般解

cos (x) = - 0.75487 \dots

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

x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

すべての解を組み合わせる

x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

10進法形式で解を証明する

x = 2.42626 \dots + 2 πn, x = - 2.42626 \dots + 2 πn

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