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人気のある三角関数 >

sin^2(x)+sin^6(x)=3cos^2(2x)

解

sin^{2} (x) + sin^{6} (x) = 3 cos^{2} (2 x)

解

x = 0.60819 \dots + 2 πn, x = π - 0.60819 \dots + 2 πn, x = - 0.60819 \dots + 2 πn, x = π + 0.60819 \dots + 2 πn, x = 1.17152 \dots + 2 πn, x = π - 1.17152 \dots + 2 πn, x = - 1.17152 \dots + 2 πn, x = π + 1.17152 \dots + 2 πn

+1

度

x = 34.84715 \dots^{\circ} + 36 0^{\circ} n, x = 145.15284 \dots^{\circ} + 36 0^{\circ} n, x = - 34.84715 \dots^{\circ} + 36 0^{\circ} n, x = 214.84715 \dots^{\circ} + 36 0^{\circ} n, x = 67.12337 \dots^{\circ} + 36 0^{\circ} n, x = 112.87662 \dots^{\circ} + 36 0^{\circ} n, x = - 67.12337 \dots^{\circ} + 36 0^{\circ} n, x = 247.12337 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sin^{2} (x) + sin^{6} (x) = 3 cos^{2} (2 x)

両辺から

3 cos^{2} (2 x)

を引く

sin^{2} (x) + sin^{6} (x) - 3 cos^{2} (2 x) = 0

三角関数の公式を使用して書き換える

sin^{2} (x) + sin^{6} (x) - 3 cos^{2} (2 x)

2倍角の公式を使用:

cos (2 x) = 1 - 2 sin^{2} (x)

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

簡素化

sin^{2} (x) + sin^{6} (x) - 3 (1 - 2 sin^{2} (x))^{2} : 13 sin^{2} (x) + sin^{6} (x) - 12 sin^{4} (x) - 3

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

(1 - 2 sin^{2} (x))^{2} : 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

完全平方式を適用する:

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

a = 1, b = 2 sin^{2} (x)

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

簡素化

1^{2} - 2 \cdot 1 \cdot 2 sin^{2} (x) + (2 sin^{2} (x))^{2} : 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

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

規則を適用

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 2 sin^{2} (x) = 4 sin^{2} (x)

2 \cdot 1 \cdot 2 sin^{2} (x)

数を乗じる:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (x)

(2 sin^{2} (x))^{2} = 4 sin^{4} (x)

(2 sin^{2} (x))^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (sin^{2} (x))^{2}

(sin^{2} (x))^{2} : sin^{4} (x)

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= sin^{2 \cdot 2} (x)

数を乗じる:

2 \cdot 2 = 4

= sin^{4} (x)

= 2^{2} sin^{4} (x)

2^{2} = 4

= 4 sin^{4} (x)

= 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

= 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

= sin^{2} (x) + sin^{6} (x) - 3 (1 - 4 sin^{2} (x) + 4 sin^{4} (x))

拡張

- 3 (1 - 4 sin^{2} (x) + 4 sin^{4} (x)) : - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

- 3 (1 - 4 sin^{2} (x) + 4 sin^{4} (x))

括弧を分配する

= (- 3) \cdot 1 + (- 3) (- 4 sin^{2} (x)) + (- 3) \cdot 4 sin^{4} (x)

マイナス・プラスの規則を適用する

+ (- a) = - a, (- a) (- b) = ab

= - 3 \cdot 1 + 3 \cdot 4 sin^{2} (x) - 3 \cdot 4 sin^{4} (x)

簡素化

- 3 \cdot 1 + 3 \cdot 4 sin^{2} (x) - 3 \cdot 4 sin^{4} (x) : - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

- 3 \cdot 1 + 3 \cdot 4 sin^{2} (x) - 3 \cdot 4 sin^{4} (x)

数を乗じる:

3 \cdot 1 = 3

= - 3 + 3 \cdot 4 sin^{2} (x) - 3 \cdot 4 sin^{4} (x)

数を乗じる:

3 \cdot 4 = 12

= - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

= - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

= sin^{2} (x) + sin^{6} (x) - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

簡素化

sin^{2} (x) + sin^{6} (x) - 3 + 12 sin^{2} (x) - 12 sin^{4} (x) : 13 sin^{2} (x) + sin^{6} (x) - 12 sin^{4} (x) - 3

sin^{2} (x) + sin^{6} (x) - 3 + 12 sin^{2} (x) - 12 sin^{4} (x)

条件のようなグループ

= sin^{2} (x) + sin^{6} (x) + 12 sin^{2} (x) - 12 sin^{4} (x) - 3

類似した元を足す:

sin^{2} (x) + 12 sin^{2} (x) = 13 sin^{2} (x)

= 13 sin^{2} (x) + sin^{6} (x) - 12 sin^{4} (x) - 3

= 13 sin^{2} (x) + sin^{6} (x) - 12 sin^{4} (x) - 3

= 13 sin^{2} (x) + sin^{6} (x) - 12 sin^{4} (x) - 3

- 3 + sin^{6} (x) - 12 sin^{4} (x) + 13 sin^{2} (x) = 0

置換で解く

- 3 + sin^{6} (x) - 12 sin^{4} (x) + 13 sin^{2} (x) = 0

仮定:

sin (x) = u

- 3 + u^{6} - 12 u^{4} + 13 u^{2} = 0

- 3 + u^{6} - 12 u^{4} + 13 u^{2} = 0 : u = 0.32648 \dots, u = - 0.32648 \dots, u = 0.84887 \dots, u = - 0.84887 \dots, u = 10.82463 \dots, u = - 10.82463 \dots

- 3 + u^{6} - 12 u^{4} + 13 u^{2} = 0

標準的な形式で書く

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

u^{6} - 12 u^{4} + 13 u^{2} - 3 = 0

equationを

v = u^{2}, v^{2} = u^{4}

と以下で書き換える:

v^{3} = u^{6}

v^{3} - 12 v^{2} + 13 v - 3 = 0

解く

v^{3} - 12 v^{2} + 13 v - 3 = 0 : v \approx 0.32648 \dots, v \approx 0.84887 \dots, v \approx 10.82463 \dots

v^{3} - 12 v^{2} + 13 v - 3 = 0

ニュートン・ラプソン法を使用して

v^{3} - 12 v^{2} + 13 v - 3 = 0

の解を1つ求める:

v \approx 0.32648 \dots

v^{3} - 12 v^{2} + 13 v - 3 = 0

ニュートン・ラプソン概算の定義

f (v) = v^{3} - 12 v^{2} + 13 v - 3

発見する

f^{^{'}} (v) : 3 v^{2} - 24 v + 13

\frac{d}{d v} (v^{3} - 12 v^{2} + 13 v - 3)

和/差の法則を適用:

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

= \frac{d}{d v} (v^{3}) - \frac{d}{d v} (12 v^{2}) + \frac{d}{d v} (13 v) - \frac{d}{d v} (3)

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

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

乗の法則を適用:

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

= 3 v^{3 - 1}

簡素化

= 3 v^{2}

\frac{d}{d v} (12 v^{2}) = 24 v

\frac{d}{d v} (12 v^{2})

定数を除去:

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

= 12 \frac{d}{d v} (v^{2})

乗の法則を適用:

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

= 12 \cdot 2 v^{2 - 1}

簡素化

= 24 v

\frac{d}{d v} (13 v) = 13

\frac{d}{d v} (13 v)

定数を除去:

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

= 13 \frac{d v}{d v}

共通の導関数を適用:

\frac{d v}{d v} = 1

= 13 \cdot 1

簡素化

= 13

\frac{d}{d v} (3) = 0

\frac{d}{d v} (3)

定数の導関数:

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

= 0

= 3 v^{2} - 24 v + 13 - 0

簡素化

= 3 v^{2} - 24 v + 13

仮定:

v_{0} = 0

Δ v_{n + 1} <

になるまで

v_{n + 1}

を計算する

0.000001

v_{1} = 0.23076 \dots : Δ v_{1} = 0.23076 \dots

f (v_{0}) = 0^{3} - 12 \cdot 0^{2} + 13 \cdot 0 - 3 = - 3

f^{^{'}} (v_{0}) = 3 \cdot 0^{2} - 24 \cdot 0 + 13 = 13

v_{1} = 0.23076 \dots

Δ v_{1} = ∣ 0.23076 \dots - 0 ∣ = 0.23076 \dots

Δ v_{1} = 0.23076 \dots

v_{2} = 0.31300 \dots : Δ v_{2} = 0.08223 \dots

f (v_{1}) = 0.23076 \dots^{3} - 12 \cdot 0.23076 \dots^{2} + 13 \cdot 0.23076 \dots - 3 = - 0.62676 \dots

f^{^{'}} (v_{1}) = 3 \cdot 0.23076 \dots^{2} - 24 \cdot 0.23076 \dots + 13 = 7.62130 \dots

v_{2} = 0.31300 \dots

Δ v_{2} = ∣ 0.31300 \dots - 0.23076 \dots ∣ = 0.08223 \dots

Δ v_{2} = 0.08223 \dots

v_{3} = 0.32613 \dots : Δ v_{3} = 0.01313 \dots

f (v_{2}) = 0.31300 \dots^{3} - 12 \cdot 0.31300 \dots^{2} + 13 \cdot 0.31300 \dots - 3 = - 0.07591 \dots

f^{^{'}} (v_{2}) = 3 \cdot 0.31300 \dots^{2} - 24 \cdot 0.31300 \dots + 13 = 5.78173 \dots

v_{3} = 0.32613 \dots

Δ v_{3} = ∣ 0.32613 \dots - 0.31300 \dots ∣ = 0.01313 \dots

Δ v_{3} = 0.01313 \dots

v_{4} = 0.32648 \dots : Δ v_{4} = 0.00034 \dots

f (v_{3}) = 0.32613 \dots^{3} - 12 \cdot 0.32613 \dots^{2} + 13 \cdot 0.32613 \dots - 3 = - 0.00190 \dots

f^{^{'}} (v_{3}) = 3 \cdot 0.32613 \dots^{2} - 24 \cdot 0.32613 \dots + 13 = 5.49177 \dots

v_{4} = 0.32648 \dots

Δ v_{4} = ∣ 0.32648 \dots - 0.32613 \dots ∣ = 0.00034 \dots

Δ v_{4} = 0.00034 \dots

v_{5} = 0.32648 \dots : Δ v_{5} = 2.41787 E - 7

f (v_{4}) = 0.32648 \dots^{3} - 12 \cdot 0.32648 \dots^{2} + 13 \cdot 0.32648 \dots - 3 = - 1.32599 E - 6

f^{^{'}} (v_{4}) = 3 \cdot 0.32648 \dots^{2} - 24 \cdot 0.32648 \dots + 13 = 5.48412 \dots

v_{5} = 0.32648 \dots

Δ v_{5} = ∣ 0.32648 \dots - 0.32648 \dots ∣ = 2.41787 E - 7

Δ v_{5} = 2.41787 E - 7

v \approx 0.32648 \dots

長除法を適用する:

\frac{v ^{3} - 12 v ^{2} + 13 v - 3}{v - 0.32648 \dots} = v^{2} - 11.67351 \dots v + 9.18876 \dots

v^{2} - 11.67351 \dots v + 9.18876 \dots \approx 0

ニュートン・ラプソン法を使用して

v^{2} - 11.67351 \dots v + 9.18876 \dots = 0

の解を1つ求める:

v \approx 0.84887 \dots

v^{2} - 11.67351 \dots v + 9.18876 \dots = 0

ニュートン・ラプソン概算の定義

f (v) = v^{2} - 11.67351 \dots v + 9.18876 \dots

発見する

f^{^{'}} (v) : 2 v - 11.67351 \dots

\frac{d}{d v} (v^{2} - 11.67351 \dots v + 9.18876 \dots)

和/差の法則を適用:

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

= \frac{d}{d v} (v^{2}) - \frac{d}{d v} (11.67351 \dots v) + \frac{d}{d v} (9.18876 \dots)

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

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

乗の法則を適用:

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

= 2 v^{2 - 1}

簡素化

= 2 v

\frac{d}{d v} (11.67351 \dots v) = 11.67351 \dots

\frac{d}{d v} (11.67351 \dots v)

定数を除去:

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

= 11.67351 \dots \frac{d v}{d v}

共通の導関数を適用:

\frac{d v}{d v} = 1

= 11.67351 \dots \cdot 1

簡素化

= 11.67351 \dots

\frac{d}{d v} (9.18876 \dots) = 0

\frac{d}{d v} (9.18876 \dots)

定数の導関数:

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

= 0

= 2 v - 11.67351 \dots + 0

簡素化

= 2 v - 11.67351 \dots

仮定:

v_{0} = 1

Δ v_{n + 1} <

になるまで

v_{n + 1}

を計算する

0.000001

v_{1} = 0.84651 \dots : Δ v_{1} = 0.15348 \dots

f (v_{0}) = 1^{2} - 11.67351 \dots \cdot 1 + 9.18876 \dots = - 1.48474 \dots

f^{^{'}} (v_{0}) = 2 \cdot 1 - 11.67351 \dots = - 9.67351 \dots

v_{1} = 0.84651 \dots

Δ v_{1} = ∣ 0.84651 \dots - 1 ∣ = 0.15348 \dots

Δ v_{1} = 0.15348 \dots

v_{2} = 0.84887 \dots : Δ v_{2} = 0.00236 \dots

f (v_{1}) = 0.84651 \dots^{2} - 11.67351 \dots \cdot 0.84651 \dots + 9.18876 \dots = 0.02355 \dots

f^{^{'}} (v_{1}) = 2 \cdot 0.84651 \dots - 11.67351 \dots = - 9.98048 \dots

v_{2} = 0.84887 \dots

Δ v_{2} = ∣ 0.84887 \dots - 0.84651 \dots ∣ = 0.00236 \dots

Δ v_{2} = 0.00236 \dots

v_{3} = 0.84887 \dots : Δ v_{3} = 5.58503 E - 7

f (v_{2}) = 0.84887 \dots^{2} - 11.67351 \dots \cdot 0.84887 \dots + 9.18876 \dots = 5.5715 E - 6

f^{^{'}} (v_{2}) = 2 \cdot 0.84887 \dots - 11.67351 \dots = - 9.97576 \dots

v_{3} = 0.84887 \dots

Δ v_{3} = ∣ 0.84887 \dots - 0.84887 \dots ∣ = 5.58503 E - 7

Δ v_{3} = 5.58503 E - 7

v \approx 0.84887 \dots

長除法を適用する:

\frac{v ^{2} - 11.67351 \dots v + 9.18876 \dots}{v - 0.84887 \dots} = v - 10.82463 \dots

v - 10.82463 \dots \approx 0

v \approx 10.82463 \dots

解答は

v \approx 0.32648 \dots, v \approx 0.84887 \dots, v \approx 10.82463 \dots

v \approx 0.32648 \dots, v \approx 0.84887 \dots, v \approx 10.82463 \dots

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 0.32648 \dots : u = 0.32648 \dots, u = - 0.32648 \dots

u^{2} = 0.32648 \dots

x^{2} = f (a)

の場合, 解は

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

u = 0.32648 \dots, u = - 0.32648 \dots

解く

u^{2} = 0.84887 \dots : u = 0.84887 \dots, u = - 0.84887 \dots

u^{2} = 0.84887 \dots

x^{2} = f (a)

の場合, 解は

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

u = 0.84887 \dots, u = - 0.84887 \dots

解く

u^{2} = 10.82463 \dots : u = 10.82463 \dots, u = - 10.82463 \dots

u^{2} = 10.82463 \dots

x^{2} = f (a)

の場合, 解は

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

u = 10.82463 \dots, u = - 10.82463 \dots

解答は

u = 0.32648 \dots, u = - 0.32648 \dots, u = 0.84887 \dots, u = - 0.84887 \dots, u = 10.82463 \dots, u = - 10.82463 \dots

代用を戻す

u = sin (x)

sin (x) = 0.32648 \dots, sin (x) = - 0.32648 \dots, sin (x) = 0.84887 \dots, sin (x) = - 0.84887 \dots, sin (x) = 10.82463 \dots, sin (x) = - 10.82463 \dots

sin (x) = 0.32648 \dots, sin (x) = - 0.32648 \dots, sin (x) = 0.84887 \dots, sin (x) = - 0.84887 \dots, sin (x) = 10.82463 \dots, sin (x) = - 10.82463 \dots

sin (x) = 0.32648 \dots : x = arcsin (0.32648 \dots) + 2 πn, x = π - arcsin (0.32648 \dots) + 2 πn

sin (x) = 0.32648 \dots

三角関数の逆数プロパティを適用する

sin (x) = 0.32648 \dots

以下の一般解

sin (x) = 0.32648 \dots

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

x = arcsin (0.32648 \dots) + 2 πn, x = π - arcsin (0.32648 \dots) + 2 πn

x = arcsin (0.32648 \dots) + 2 πn, x = π - arcsin (0.32648 \dots) + 2 πn

sin (x) = - 0.32648 \dots : x = arcsin (- 0.32648 \dots) + 2 πn, x = π + arcsin (0.32648 \dots) + 2 πn

sin (x) = - 0.32648 \dots

三角関数の逆数プロパティを適用する

sin (x) = - 0.32648 \dots

以下の一般解

sin (x) = - 0.32648 \dots

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

x = arcsin (- 0.32648 \dots) + 2 πn, x = π + arcsin (0.32648 \dots) + 2 πn

x = arcsin (- 0.32648 \dots) + 2 πn, x = π + arcsin (0.32648 \dots) + 2 πn

sin (x) = 0.84887 \dots : x = arcsin (0.84887 \dots) + 2 πn, x = π - arcsin (0.84887 \dots) + 2 πn

sin (x) = 0.84887 \dots

三角関数の逆数プロパティを適用する

sin (x) = 0.84887 \dots

以下の一般解

sin (x) = 0.84887 \dots

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

x = arcsin (0.84887 \dots) + 2 πn, x = π - arcsin (0.84887 \dots) + 2 πn

x = arcsin (0.84887 \dots) + 2 πn, x = π - arcsin (0.84887 \dots) + 2 πn

sin (x) = - 0.84887 \dots : x = arcsin (- 0.84887 \dots) + 2 πn, x = π + arcsin (0.84887 \dots) + 2 πn

sin (x) = - 0.84887 \dots

三角関数の逆数プロパティを適用する

sin (x) = - 0.84887 \dots

以下の一般解

sin (x) = - 0.84887 \dots

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

x = arcsin (- 0.84887 \dots) + 2 πn, x = π + arcsin (0.84887 \dots) + 2 πn

x = arcsin (- 0.84887 \dots) + 2 πn, x = π + arcsin (0.84887 \dots) + 2 πn

sin (x) = 10.82463 \dots :

解なし

sin (x) = 10.82463 \dots

- 1 \leq sin (x) \leq 1

解なし

sin (x) = - 10.82463 \dots :

解なし

sin (x) = - 10.82463 \dots

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (0.32648 \dots) + 2 πn, x = π - arcsin (0.32648 \dots) + 2 πn, x = arcsin (- 0.32648 \dots) + 2 πn, x = π + arcsin (0.32648 \dots) + 2 πn, x = arcsin (0.84887 \dots) + 2 πn, x = π - arcsin (0.84887 \dots) + 2 πn, x = arcsin (- 0.84887 \dots) + 2 πn, x = π + arcsin (0.84887 \dots) + 2 πn

10進法形式で解を証明する

x = 0.60819 \dots + 2 πn, x = π - 0.60819 \dots + 2 πn, x = - 0.60819 \dots + 2 πn, x = π + 0.60819 \dots + 2 πn, x = 1.17152 \dots + 2 πn, x = π - 1.17152 \dots + 2 πn, x = - 1.17152 \dots + 2 πn, x = π + 1.17152 \dots + 2 πn

グラフ

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