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人気のある三角関数 >

sin^3(x)+sin(x)-4=0

解

sin^{3} (x) + sin (x) - 4 = 0

解

以下の解はない : x \in R

解答ステップ

sin^{3} (x) + sin (x) - 4 = 0

置換で解く

sin^{3} (x) + sin (x) - 4 = 0

仮定:

sin (x) = u

u^{3} + u - 4 = 0

u^{3} + u - 4 = 0 : u \approx 1.37879 \dots

u^{3} + u - 4 = 0

ニュートン・ラプソン法を使用して

u^{3} + u - 4 = 0

の解を1つ求める:

u \approx 1.37879 \dots

u^{3} + u - 4 = 0

ニュートン・ラプソン概算の定義

f (u) = u^{3} + u - 4

発見する

f^{^{'}} (u) : 3 u^{2} + 1

\frac{d}{d u} (u^{3} + u - 4)

和/差の法則を適用:

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

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

\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 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} (4) = 0

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

定数の導関数:

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

= 0

= 3 u^{2} + 1 - 0

簡素化

= 3 u^{2} + 1

仮定:

u_{0} = 4

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = 2.69387 \dots : Δ u_{1} = 1.30612 \dots

f (u_{0}) = 4^{3} + 4 - 4 = 64

f^{^{'}} (u_{0}) = 3 \cdot 4^{2} + 1 = 49

u_{1} = 2.69387 \dots

Δ u_{1} = ∣ 2.69387 \dots - 4 ∣ = 1.30612 \dots

Δ u_{1} = 1.30612 \dots

u_{2} = 1.89271 \dots : Δ u_{2} = 0.80116 \dots

f (u_{1}) = 2.69387 \dots^{3} + 2.69387 \dots - 4 = 18.24328 \dots

f^{^{'}} (u_{1}) = 3 \cdot 2.69387 \dots^{2} + 1 = 22.77092 \dots

u_{2} = 1.89271 \dots

Δ u_{2} = ∣ 1.89271 \dots - 2.69387 \dots ∣ = 0.80116 \dots

Δ u_{2} = 0.80116 \dots

u_{3} = 1.49490 \dots : Δ u_{3} = 0.39780 \dots

f (u_{2}) = 1.89271 \dots^{3} + 1.89271 \dots - 4 = 4.67308 \dots

f^{^{'}} (u_{2}) = 3 \cdot 1.89271 \dots^{2} + 1 = 11.74707 \dots

u_{3} = 1.49490 \dots

Δ u_{3} = ∣ 1.49490 \dots - 1.89271 \dots ∣ = 0.39780 \dots

Δ u_{3} = 0.39780 \dots

u_{4} = 1.38644 \dots : Δ u_{4} = 0.10846 \dots

f (u_{3}) = 1.49490 \dots^{3} + 1.49490 \dots - 4 = 0.83561 \dots

f^{^{'}} (u_{3}) = 3 \cdot 1.49490 \dots^{2} + 1 = 7.70421 \dots

u_{4} = 1.38644 \dots

Δ u_{4} = ∣ 1.38644 \dots - 1.49490 \dots ∣ = 0.10846 \dots

Δ u_{4} = 0.10846 \dots

u_{5} = 1.37883 \dots : Δ u_{5} = 0.00760 \dots

f (u_{4}) = 1.38644 \dots^{3} + 1.38644 \dots - 4 = 0.05148 \dots

f^{^{'}} (u_{4}) = 3 \cdot 1.38644 \dots^{2} + 1 = 6.76665 \dots

u_{5} = 1.37883 \dots

Δ u_{5} = ∣ 1.37883 \dots - 1.38644 \dots ∣ = 0.00760 \dots

Δ u_{5} = 0.00760 \dots

u_{6} = 1.37879 \dots : Δ u_{6} = 0.00003 \dots

f (u_{5}) = 1.37883 \dots^{3} + 1.37883 \dots - 4 = 0.00024 \dots

f^{^{'}} (u_{5}) = 3 \cdot 1.37883 \dots^{2} + 1 = 6.70353 \dots

u_{6} = 1.37879 \dots

Δ u_{6} = ∣ 1.37879 \dots - 1.37883 \dots ∣ = 0.00003 \dots

Δ u_{6} = 0.00003 \dots

u_{7} = 1.37879 \dots : Δ u_{7} = 7.93136 E - 10

f (u_{6}) = 1.37879 \dots^{3} + 1.37879 \dots - 4 = 5.31658 E - 9

f^{^{'}} (u_{6}) = 3 \cdot 1.37879 \dots^{2} + 1 = 6.70324 \dots

u_{7} = 1.37879 \dots

Δ u_{7} = ∣ 1.37879 \dots - 1.37879 \dots ∣ = 7.93136 E - 10

Δ u_{7} = 7.93136 E - 10

u \approx 1.37879 \dots

長除法を適用する:

\frac{u ^{3} + u - 4}{u - 1.37879 \dots} = u^{2} + 1.37879 \dots u + 2.90108 \dots

u^{2} + 1.37879 \dots u + 2.90108 \dots \approx 0

ニュートン・ラプソン法を使用して

u^{2} + 1.37879 \dots u + 2.90108 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

u^{2} + 1.37879 \dots u + 2.90108 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = u^{2} + 1.37879 \dots u + 2.90108 \dots

発見する

f^{^{'}} (u) : 2 u + 1.37879 \dots

\frac{d}{d u} (u^{2} + 1.37879 \dots u + 2.90108 \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} (1.37879 \dots u) + \frac{d}{d u} (2.90108 \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} (1.37879 \dots u) = 1.37879 \dots

\frac{d}{d u} (1.37879 \dots u)

定数を除去:

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

= 1.37879 \dots \frac{d u}{d u}

共通の導関数を適用:

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

= 1.37879 \dots \cdot 1

簡素化

= 1.37879 \dots

\frac{d}{d u} (2.90108 \dots) = 0

\frac{d}{d u} (2.90108 \dots)

定数の導関数:

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

= 0

= 2 u + 1.37879 \dots + 0

簡素化

= 2 u + 1.37879 \dots

仮定:

u_{0} = - 2

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = - 0.41924 \dots : Δ u_{1} = 1.58075 \dots

f (u_{0}) = (- 2)^{2} + 1.37879 \dots (- 2) + 2.90108 \dots = 4.14348 \dots

f^{^{'}} (u_{0}) = 2 (- 2) + 1.37879 \dots = - 2.62120 \dots

u_{1} = - 0.41924 \dots

Δ u_{1} = ∣ - 0.41924 \dots - (- 2) ∣ = 1.58075 \dots

Δ u_{1} = 1.58075 \dots

u_{2} = - 5.04396 \dots : Δ u_{2} = 4.62472 \dots

f (u_{1}) = (- 0.41924 \dots)^{2} + 1.37879 \dots (- 0.41924 \dots) + 2.90108 \dots = 2.49879 \dots

f^{^{'}} (u_{1}) = 2 (- 0.41924 \dots) + 1.37879 \dots = 0.54031 \dots

u_{2} = - 5.04396 \dots

Δ u_{2} = ∣ - 5.04396 \dots - (- 0.41924 \dots) ∣ = 4.62472 \dots

Δ u_{2} = 4.62472 \dots

u_{3} = - 2.58814 \dots : Δ u_{3} = 2.45582 \dots

f (u_{2}) = (- 5.04396 \dots)^{2} + 1.37879 \dots (- 5.04396 \dots) + 2.90108 \dots = 21.38809 \dots

f^{^{'}} (u_{2}) = 2 (- 5.04396 \dots) + 1.37879 \dots = - 8.70914 \dots

u_{3} = - 2.58814 \dots

Δ u_{3} = ∣ - 2.58814 \dots - (- 5.04396 \dots) ∣ = 2.45582 \dots

Δ u_{3} = 2.45582 \dots

u_{4} = - 0.99998 \dots : Δ u_{4} = 1.58816 \dots

f (u_{3}) = (- 2.58814 \dots)^{2} + 1.37879 \dots (- 2.58814 \dots) + 2.90108 \dots = 6.03105 \dots

f^{^{'}} (u_{3}) = 2 (- 2.58814 \dots) + 1.37879 \dots = - 3.79749 \dots

u_{4} = - 0.99998 \dots

Δ u_{4} = ∣ - 0.99998 \dots - (- 2.58814 \dots) ∣ = 1.58816 \dots

Δ u_{4} = 1.58816 \dots

u_{5} = 3.06056 \dots : Δ u_{5} = 4.06054 \dots

f (u_{4}) = (- 0.99998 \dots)^{2} + 1.37879 \dots (- 0.99998 \dots) + 2.90108 \dots = 2.52227 \dots

f^{^{'}} (u_{4}) = 2 (- 0.99998 \dots) + 1.37879 \dots = - 0.62116 \dots

u_{5} = 3.06056 \dots

Δ u_{5} = ∣ 3.06056 \dots - (- 0.99998 \dots) ∣ = 4.06054 \dots

Δ u_{5} = 4.06054 \dots

u_{6} = 0.86213 \dots : Δ u_{6} = 2.19842 \dots

f (u_{5}) = 3.06056 \dots^{2} + 1.37879 \dots \cdot 3.06056 \dots + 2.90108 \dots = 16.48802 \dots

f^{^{'}} (u_{5}) = 2 \cdot 3.06056 \dots + 1.37879 \dots = 7.49992 \dots

u_{6} = 0.86213 \dots

Δ u_{6} = ∣ 0.86213 \dots - 3.06056 \dots ∣ = 2.19842 \dots

Δ u_{6} = 2.19842 \dots

u_{7} = - 0.69537 \dots : Δ u_{7} = 1.55751 \dots

f (u_{6}) = 0.86213 \dots^{2} + 1.37879 \dots \cdot 0.86213 \dots + 2.90108 \dots = 4.83307 \dots

f^{^{'}} (u_{6}) = 2 \cdot 0.86213 \dots + 1.37879 \dots = 3.10307 \dots

u_{7} = - 0.69537 \dots

Δ u_{7} = ∣ - 0.69537 \dots - 0.86213 \dots ∣ = 1.55751 \dots

Δ u_{7} = 1.55751 \dots

u_{8} = 202.24500 \dots : Δ u_{8} = 202.94037 \dots

f (u_{7}) = (- 0.69537 \dots)^{2} + 1.37879 \dots (- 0.69537 \dots) + 2.90108 \dots = 2.42584 \dots

f^{^{'}} (u_{7}) = 2 (- 0.69537 \dots) + 1.37879 \dots = - 0.01195 \dots

u_{8} = 202.24500 \dots

Δ u_{8} = ∣ 202.24500 \dots - (- 0.69537 \dots) ∣ = 202.94037 \dots

Δ u_{8} = 202.94037 \dots

u_{9} = 100.77182 \dots : Δ u_{9} = 101.47317 \dots

f (u_{8}) = 202.24500 \dots^{2} + 1.37879 \dots \cdot 202.24500 \dots + 2.90108 \dots = 41184.79638 \dots

f^{^{'}} (u_{8}) = 2 \cdot 202.24500 \dots + 1.37879 \dots = 405.86879 \dots

u_{9} = 100.77182 \dots

Δ u_{9} = ∣ 100.77182 \dots - 202.24500 \dots ∣ = 101.47317 \dots

Δ u_{9} = 101.47317 \dots

u_{10} = 50.02925 \dots : Δ u_{10} = 50.74256 \dots

f (u_{9}) = 100.77182 \dots^{2} + 1.37879 \dots \cdot 100.77182 \dots + 2.90108 \dots = 10296.80558 \dots

f^{^{'}} (u_{9}) = 2 \cdot 100.77182 \dots + 1.37879 \dots = 202.92244 \dots

u_{10} = 50.02925 \dots

Δ u_{10} = ∣ 50.02925 \dots - 100.77182 \dots ∣ = 50.74256 \dots

Δ u_{10} = 50.74256 \dots

u_{11} = 24.64601 \dots : Δ u_{11} = 25.38324 \dots

f (u_{10}) = 50.02925 \dots^{2} + 1.37879 \dots \cdot 50.02925 \dots + 2.90108 \dots = 2574.80799 \dots

f^{^{'}} (u_{10}) = 2 \cdot 50.02925 \dots + 1.37879 \dots = 101.43731 \dots

u_{11} = 24.64601 \dots

Δ u_{11} = ∣ 24.64601 \dots - 50.02925 \dots ∣ = 25.38324 \dots

Δ u_{11} = 25.38324 \dots

u_{12} = 11.93043 \dots : Δ u_{12} = 12.71558 \dots

f (u_{11}) = 24.64601 \dots^{2} + 1.37879 \dots \cdot 24.64601 \dots + 2.90108 \dots = 644.30902 \dots

f^{^{'}} (u_{11}) = 2 \cdot 24.64601 \dots + 1.37879 \dots = 50.67082 \dots

u_{12} = 11.93043 \dots

Δ u_{12} = ∣ 11.93043 \dots - 24.64601 \dots ∣ = 12.71558 \dots

Δ u_{12} = 12.71558 \dots

u_{13} = 5.52440 \dots : Δ u_{13} = 6.40602 \dots

f (u_{12}) = 11.93043 \dots^{2} + 1.37879 \dots \cdot 11.93043 \dots + 2.90108 \dots = 161.68600 \dots

f^{^{'}} (u_{12}) = 2 \cdot 11.93043 \dots + 1.37879 \dots = 25.23966 \dots

u_{13} = 5.52440 \dots

Δ u_{13} = ∣ 5.52440 \dots - 11.93043 \dots ∣ = 6.40602 \dots

Δ u_{13} = 6.40602 \dots

u_{14} = 2.22230 \dots : Δ u_{14} = 3.30209 \dots

f (u_{13}) = 5.52440 \dots^{2} + 1.37879 \dots \cdot 5.52440 \dots + 2.90108 \dots = 41.03718 \dots

f^{^{'}} (u_{13}) = 2 \cdot 5.52440 \dots + 1.37879 \dots = 12.42761 \dots

u_{14} = 2.22230 \dots

Δ u_{14} = ∣ 2.22230 \dots - 5.52440 \dots ∣ = 3.30209 \dots

Δ u_{14} = 3.30209 \dots

u_{15} = 0.34989 \dots : Δ u_{15} = 1.87241 \dots

f (u_{14}) = 2.22230 \dots^{2} + 1.37879 \dots \cdot 2.22230 \dots + 2.90108 \dots = 10.90385 \dots

f^{^{'}} (u_{14}) = 2 \cdot 2.22230 \dots + 1.37879 \dots = 5.82341 \dots

u_{15} = 0.34989 \dots

Δ u_{15} = ∣ 0.34989 \dots - 2.22230 \dots ∣ = 1.87241 \dots

Δ u_{15} = 1.87241 \dots

u_{16} = - 1.33680 \dots : Δ u_{16} = 1.68669 \dots

f (u_{15}) = 0.34989 \dots^{2} + 1.37879 \dots \cdot 0.34989 \dots + 2.90108 \dots = 3.50593 \dots

f^{^{'}} (u_{15}) = 2 \cdot 0.34989 \dots + 1.37879 \dots = 2.07858 \dots

u_{16} = - 1.33680 \dots

Δ u_{16} = ∣ - 1.33680 \dots - 0.34989 \dots ∣ = 1.68669 \dots

Δ u_{16} = 1.68669 \dots

u_{17} = 0.86039 \dots : Δ u_{17} = 2.19719 \dots

f (u_{16}) = (- 1.33680 \dots)^{2} + 1.37879 \dots (- 1.33680 \dots) + 2.90108 \dots = 2.84494 \dots

f^{^{'}} (u_{16}) = 2 (- 1.33680 \dots) + 1.37879 \dots = - 1.29480 \dots

u_{17} = 0.86039 \dots

Δ u_{17} = ∣ 0.86039 \dots - (- 1.33680 \dots) ∣ = 2.19719 \dots

Δ u_{17} = 2.19719 \dots

解を見つけられない

解は

u \approx 1.37879 \dots

代用を戻す

u = sin (x)

sin (x) \approx 1.37879 \dots

sin (x) \approx 1.37879 \dots

sin (x) = 1.37879 \dots :

解なし

sin (x) = 1.37879 \dots

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

以下の解はない : x \in R

グラフ

人気の例

solvefor a,sin^2(a)+cos^2(b)=1

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