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人気のある三角関数 >

sin^5(x)+sin(x)+2sin^2(x)=1

解

sin^{5} (x) + sin (x) + 2 sin^{2} (x) = 1

解

x = 0.51263 \dots + 2 πn, x = π - 0.51263 \dots + 2 πn

+1

度

x = 29.37198 \dots^{\circ} + 36 0^{\circ} n, x = 150.62801 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sin^{5} (x) + sin (x) + 2 sin^{2} (x) = 1

置換で解く

sin^{5} (x) + sin (x) + 2 sin^{2} (x) = 1

仮定:

sin (x) = u

u^{5} + u + 2 u^{2} = 1

u^{5} + u + 2 u^{2} = 1 : u \approx 0.49047 \dots

u^{5} + u + 2 u^{2} = 1

1

を左側に移動します

u^{5} + u + 2 u^{2} = 1

両辺から

1

を引く

u^{5} + u + 2 u^{2} - 1 = 1 - 1

簡素化

u^{5} + u + 2 u^{2} - 1 = 0

u^{5} + u + 2 u^{2} - 1 = 0

標準的な形式で書く

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

u^{5} + 2 u^{2} + u - 1 = 0

ニュートン・ラプソン法を使用して

u^{5} + 2 u^{2} + u - 1 = 0

の解を1つ求める:

u \approx 0.49047 \dots

u^{5} + 2 u^{2} + u - 1 = 0

ニュートン・ラプソン概算の定義

f (u) = u^{5} + 2 u^{2} + u - 1

発見する

f^{^{'}} (u) : 5 u^{4} + 4 u + 1

\frac{d}{d u} (u^{5} + 2 u^{2} + u - 1)

和/差の法則を適用:

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

= \frac{d}{d u} (u^{5}) + \frac{d}{d u} (2 u^{2}) + \frac{d u}{d u} - \frac{d}{d u} (1)

\frac{d}{d u} (u^{5}) = 5 u^{4}

\frac{d}{d u} (u^{5})

乗の法則を適用:

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

= 5 u^{5 - 1}

簡素化

= 5 u^{4}

\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} (1) = 0

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

定数の導関数:

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

= 0

= 5 u^{4} + 4 u + 1 - 0

簡素化

= 5 u^{4} + 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} = 0.7 : Δ u_{1} = 0.3

f (u_{0}) = 1^{5} + 2 \cdot 1^{2} + 1 - 1 = 3

f^{^{'}} (u_{0}) = 5 \cdot 1^{4} + 4 \cdot 1 + 1 = 10

u_{1} = 0.7

Δ u_{1} = ∣ 0.7 - 1 ∣ = 0.3

Δ u_{1} = 0.3

u_{2} = 0.53040 \dots : Δ u_{2} = 0.16959 \dots

f (u_{1}) = 0. 7^{5} + 2 \cdot 0. 7^{2} + 0.7 - 1 = 0.84807

f^{^{'}} (u_{1}) = 5 \cdot 0. 7^{4} + 4 \cdot 0.7 + 1 = 5.0005

u_{2} = 0.53040 \dots

Δ u_{2} = ∣ 0.53040 \dots - 0.7 ∣ = 0.16959 \dots

Δ u_{2} = 0.16959 \dots

u_{3} = 0.49201 \dots : Δ u_{3} = 0.03839 \dots

f (u_{2}) = 0.53040 \dots^{5} + 2 \cdot 0.53040 \dots^{2} + 0.53040 \dots - 1 = 0.13503 \dots

f^{^{'}} (u_{2}) = 5 \cdot 0.53040 \dots^{4} + 4 \cdot 0.53040 \dots + 1 = 3.51733 \dots

u_{3} = 0.49201 \dots

Δ u_{3} = ∣ 0.49201 \dots - 0.53040 \dots ∣ = 0.03839 \dots

Δ u_{3} = 0.03839 \dots

u_{4} = 0.49048 \dots : Δ u_{4} = 0.00153 \dots

f (u_{3}) = 0.49201 \dots^{5} + 2 \cdot 0.49201 \dots^{2} + 0.49201 \dots - 1 = 0.00499 \dots

f^{^{'}} (u_{3}) = 5 \cdot 0.49201 \dots^{4} + 4 \cdot 0.49201 \dots + 1 = 3.26104 \dots

u_{4} = 0.49048 \dots

Δ u_{4} = ∣ 0.49048 \dots - 0.49201 \dots ∣ = 0.00153 \dots

Δ u_{4} = 0.00153 \dots

u_{5} = 0.49047 \dots : Δ u_{5} = 2.29877 E - 6

f (u_{4}) = 0.49048 \dots^{5} + 2 \cdot 0.49048 \dots^{2} + 0.49048 \dots - 1 = 7.47396 E - 6

f^{^{'}} (u_{4}) = 5 \cdot 0.49048 \dots^{4} + 4 \cdot 0.49048 \dots + 1 = 3.25129 \dots

u_{5} = 0.49047 \dots

Δ u_{5} = ∣ 0.49047 \dots - 0.49048 \dots ∣ = 2.29877 E - 6

Δ u_{5} = 2.29877 E - 6

u_{6} = 0.49047 \dots : Δ u_{6} = 5.1684 E - 12

f (u_{5}) = 0.49047 \dots^{5} + 2 \cdot 0.49047 \dots^{2} + 0.49047 \dots - 1 = 1.68039 E - 11

f^{^{'}} (u_{5}) = 5 \cdot 0.49047 \dots^{4} + 4 \cdot 0.49047 \dots + 1 = 3.25127 \dots

u_{6} = 0.49047 \dots

Δ u_{6} = ∣ 0.49047 \dots - 0.49047 \dots ∣ = 5.1684 E - 12

Δ u_{6} = 5.1684 E - 12

u \approx 0.49047 \dots

長除法を適用する:

\frac{u ^{5} + 2 u ^{2} + u - 1}{u - 0.49047 \dots} = u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots

u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots \approx 0

ニュートン・ラプソン法を使用して

u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots

発見する

f^{^{'}} (u) : 4 u^{3} + 1.47143 \dots u^{2} + 0.48113 \dots u + 2.11799 \dots

\frac{d}{d u} (u^{4} + 0.49047 \dots u^{3} + 0.24056 \dots u^{2} + 2.11799 \dots u + 2.03882 \dots)

和/差の法則を適用:

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

= \frac{d}{d u} (u^{4}) + \frac{d}{d u} (0.49047 \dots u^{3}) + \frac{d}{d u} (0.24056 \dots u^{2}) + \frac{d}{d u} (2.11799 \dots u) + \frac{d}{d u} (2.03882 \dots)

\frac{d}{d u} (u^{4}) = 4 u^{3}

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

乗の法則を適用:

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

= 4 u^{4 - 1}

簡素化

= 4 u^{3}

\frac{d}{d u} (0.49047 \dots u^{3}) = 1.47143 \dots u^{2}

\frac{d}{d u} (0.49047 \dots u^{3})

定数を除去:

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

= 0.49047 \dots \frac{d}{d u} (u^{3})

乗の法則を適用:

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

= 0.49047 \dots \cdot 3 u^{3 - 1}

簡素化

= 1.47143 \dots u^{2}

\frac{d}{d u} (0.24056 \dots u^{2}) = 0.48113 \dots u

\frac{d}{d u} (0.24056 \dots u^{2})

定数を除去:

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

= 0.24056 \dots \frac{d}{d u} (u^{2})

乗の法則を適用:

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

= 0.24056 \dots \cdot 2 u^{2 - 1}

簡素化

= 0.48113 \dots u

\frac{d}{d u} (2.11799 \dots u) = 2.11799 \dots

\frac{d}{d u} (2.11799 \dots u)

定数を除去:

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

= 2.11799 \dots \frac{d u}{d u}

共通の導関数を適用:

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

= 2.11799 \dots \cdot 1

簡素化

= 2.11799 \dots

\frac{d}{d u} (2.03882 \dots) = 0

\frac{d}{d u} (2.03882 \dots)

定数の導関数:

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

= 0

= 4 u^{3} + 1.47143 \dots u^{2} + 0.48113 \dots u + 2.11799 \dots + 0

簡素化

= 4 u^{3} + 1.47143 \dots u^{2} + 0.48113 \dots u + 2.11799 \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} = - 0.24759 \dots : Δ u_{1} = 0.75240 \dots

f (u_{0}) = (- 1)^{4} + 0.49047 \dots (- 1)^{3} + 0.24056 \dots (- 1)^{2} + 2.11799 \dots (- 1) + 2.03882 \dots = 0.67092 \dots

f^{^{'}} (u_{0}) = 4 (- 1)^{3} + 1.47143 \dots (- 1)^{2} + 0.48113 \dots (- 1) + 2.11799 \dots = - 0.89171 \dots

u_{1} = - 0.24759 \dots

Δ u_{1} = ∣ - 0.24759 \dots - (- 1) ∣ = 0.75240 \dots

Δ u_{1} = 0.75240 \dots

u_{2} = - 0.99967 \dots : Δ u_{2} = 0.75207 \dots

f (u_{1}) = (- 0.24759 \dots)^{4} + 0.49047 \dots (- 0.24759 \dots)^{3} + 0.24056 \dots (- 0.24759 \dots)^{2} + 2.11799 \dots (- 0.24759 \dots) + 2.03882 \dots = 1.52548 \dots

f^{^{'}} (u_{1}) = 4 (- 0.24759 \dots)^{3} + 1.47143 \dots (- 0.24759 \dots)^{2} + 0.48113 \dots (- 0.24759 \dots) + 2.11799 \dots = 2.02835 \dots

u_{2} = - 0.99967 \dots

Δ u_{2} = ∣ - 0.99967 \dots - (- 0.24759 \dots) ∣ = 0.75207 \dots

Δ u_{2} = 0.75207 \dots

u_{3} = - 0.24497 \dots : Δ u_{3} = 0.75470 \dots

f (u_{2}) = (- 0.99967 \dots)^{4} + 0.49047 \dots (- 0.99967 \dots)^{3} + 0.24056 \dots (- 0.99967 \dots)^{2} + 2.11799 \dots (- 0.99967 \dots) + 2.03882 \dots = 0.67063 \dots

f^{^{'}} (u_{2}) = 4 (- 0.99967 \dots)^{3} + 1.47143 \dots (- 0.99967 \dots)^{2} + 0.48113 \dots (- 0.99967 \dots) + 2.11799 \dots = - 0.88860 \dots

u_{3} = - 0.24497 \dots

Δ u_{3} = ∣ - 0.24497 \dots - (- 0.99967 \dots) ∣ = 0.75470 \dots

Δ u_{3} = 0.75470 \dots

u_{4} = - 0.99920 \dots : Δ u_{4} = 0.75423 \dots

f (u_{3}) = (- 0.24497 \dots)^{4} + 0.49047 \dots (- 0.24497 \dots)^{3} + 0.24056 \dots (- 0.24497 \dots)^{2} + 2.11799 \dots (- 0.24497 \dots) + 2.03882 \dots = 1.53081 \dots

f^{^{'}} (u_{3}) = 4 (- 0.24497 \dots)^{3} + 1.47143 \dots (- 0.24497 \dots)^{2} + 0.48113 \dots (- 0.24497 \dots) + 2.11799 \dots = 2.02962 \dots

u_{4} = - 0.99920 \dots

Δ u_{4} = ∣ - 0.99920 \dots - (- 0.24497 \dots) ∣ = 0.75423 \dots

Δ u_{4} = 0.75423 \dots

u_{5} = - 0.24113 \dots : Δ u_{5} = 0.75806 \dots

f (u_{4}) = (- 0.99920 \dots)^{4} + 0.49047 \dots (- 0.99920 \dots)^{3} + 0.24056 \dots (- 0.99920 \dots)^{2} + 2.11799 \dots (- 0.99920 \dots) + 2.03882 \dots = 0.67021 \dots

f^{^{'}} (u_{4}) = 4 (- 0.99920 \dots)^{3} + 1.47143 \dots (- 0.99920 \dots)^{2} + 0.48113 \dots (- 0.99920 \dots) + 2.11799 \dots = - 0.88411 \dots

u_{5} = - 0.24113 \dots

Δ u_{5} = ∣ - 0.24113 \dots - (- 0.99920 \dots) ∣ = 0.75806 \dots

Δ u_{5} = 0.75806 \dots

u_{6} = - 0.99852 \dots : Δ u_{6} = 0.75739 \dots

f (u_{5}) = (- 0.24113 \dots)^{4} + 0.49047 \dots (- 0.24113 \dots)^{3} + 0.24056 \dots (- 0.24113 \dots)^{2} + 2.11799 \dots (- 0.24113 \dots) + 2.03882 \dots = 1.53860 \dots

f^{^{'}} (u_{5}) = 4 (- 0.24113 \dots)^{3} + 1.47143 \dots (- 0.24113 \dots)^{2} + 0.48113 \dots (- 0.24113 \dots) + 2.11799 \dots = 2.03144 \dots

u_{6} = - 0.99852 \dots

Δ u_{6} = ∣ - 0.99852 \dots - (- 0.24113 \dots) ∣ = 0.75739 \dots

Δ u_{6} = 0.75739 \dots

u_{7} = - 0.23556 \dots : Δ u_{7} = 0.76296 \dots

f (u_{6}) = (- 0.99852 \dots)^{4} + 0.49047 \dots (- 0.99852 \dots)^{3} + 0.24056 \dots (- 0.99852 \dots)^{2} + 2.11799 \dots (- 0.99852 \dots) + 2.03882 \dots = 0.66962 \dots

f^{^{'}} (u_{6}) = 4 (- 0.99852 \dots)^{3} + 1.47143 \dots (- 0.99852 \dots)^{2} + 0.48113 \dots (- 0.99852 \dots) + 2.11799 \dots = - 0.87765 \dots

u_{7} = - 0.23556 \dots

Δ u_{7} = ∣ - 0.23556 \dots - (- 0.99852 \dots) ∣ = 0.76296 \dots

Δ u_{7} = 0.76296 \dots

u_{8} = - 0.99756 \dots : Δ u_{8} = 0.76200 \dots

f (u_{7}) = (- 0.23556 \dots)^{4} + 0.49047 \dots (- 0.23556 \dots)^{3} + 0.24056 \dots (- 0.23556 \dots)^{2} + 2.11799 \dots (- 0.23556 \dots) + 2.03882 \dots = 1.54992 \dots

f^{^{'}} (u_{7}) = 4 (- 0.23556 \dots)^{3} + 1.47143 \dots (- 0.23556 \dots)^{2} + 0.48113 \dots (- 0.23556 \dots) + 2.11799 \dots = 2.03401 \dots

u_{8} = - 0.99756 \dots

Δ u_{8} = ∣ - 0.99756 \dots - (- 0.23556 \dots) ∣ = 0.76200 \dots

Δ u_{8} = 0.76200 \dots

u_{9} = - 0.22755 \dots : Δ u_{9} = 0.77001 \dots

f (u_{8}) = (- 0.99756 \dots)^{4} + 0.49047 \dots (- 0.99756 \dots)^{3} + 0.24056 \dots (- 0.99756 \dots)^{2} + 2.11799 \dots (- 0.99756 \dots) + 2.03882 \dots = 0.66878 \dots

f^{^{'}} (u_{8}) = 4 (- 0.99756 \dots)^{3} + 1.47143 \dots (- 0.99756 \dots)^{2} + 0.48113 \dots (- 0.99756 \dots) + 2.11799 \dots = - 0.86853 \dots

u_{9} = - 0.22755 \dots

Δ u_{9} = ∣ - 0.22755 \dots - (- 0.99756 \dots) ∣ = 0.77001 \dots

Δ u_{9} = 0.77001 \dots

解を見つけられない

解は

u \approx 0.49047 \dots

代用を戻す

u = sin (x)

sin (x) \approx 0.49047 \dots

sin (x) \approx 0.49047 \dots

sin (x) = 0.49047 \dots : x = arcsin (0.49047 \dots) + 2 πn, x = π - arcsin (0.49047 \dots) + 2 πn

sin (x) = 0.49047 \dots

三角関数の逆数プロパティを適用する

sin (x) = 0.49047 \dots

以下の一般解

sin (x) = 0.49047 \dots

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (0.49047 \dots) + 2 πn, x = π - arcsin (0.49047 \dots) + 2 πn

x = arcsin (0.49047 \dots) + 2 πn, x = π - arcsin (0.49047 \dots) + 2 πn

すべての解を組み合わせる

x = arcsin (0.49047 \dots) + 2 πn, x = π - arcsin (0.49047 \dots) + 2 πn

10進法形式で解を証明する

x = 0.51263 \dots + 2 πn, x = π - 0.51263 \dots + 2 πn

グラフ

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