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人気のある三角関数 >

sin^4(x)+sin^2(x)=sin^6(x)

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

sin (x) = u

u^{4} + u^{2} = u^{6}

u^{4} + u^{2} = u^{6} : u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{4} + u^{2} = u^{6}

辺を交換する

u^{6} = u^{4} + u^{2}

u^{2}

を左側に移動します

u^{6} = u^{4} + u^{2}

両辺から

u^{2}

を引く

u^{6} - u^{2} = u^{4} + u^{2} - u^{2}

簡素化

u^{6} - u^{2} = u^{4}

u^{6} - u^{2} = u^{4}

u^{4}

を左側に移動します

u^{6} - u^{2} = u^{4}

両辺から

u^{4}

を引く

u^{6} - u^{2} - u^{4} = u^{4} - u^{4}

簡素化

u^{6} - u^{2} - u^{4} = 0

u^{6} - u^{2} - u^{4} = 0

標準的な形式で書く

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

u^{6} - u^{4} - u^{2} = 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} - v^{2} - v = 0

解く

v^{3} - v^{2} - v = 0 : v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{3} - v^{2} - v = 0

因数

v^{3} - v^{2} - v : v (v^{2} - v - 1)

v^{3} - v^{2} - v

指数の規則を適用する:

a^{b + c} = a^{b} a^{c}

v^{2} = vv

= v^{2} v - vv - v

共通項をくくり出す

v

= v (v^{2} - v - 1)

v (v^{2} - v - 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

v = 0 or v^{2} - v - 1 = 0

解く

v^{2} - v - 1 = 0 : v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{2} - v - 1 = 0

解くとthe二次式

v^{2} - v - 1 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 1, c = - 1

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

(- 1)^{2} - 4 \cdot 1 \cdot (- 1)

規則を適用

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 1)^{2} = 1^{2}

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

解を分離する

v_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

v = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

v = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

\frac{- ( - 1 ) - 5}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次equationの解:

v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

解答は

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

解く

u^{2} = \frac{1 + 5}{2} : u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

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

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

解く

u^{2} = \frac{1 - 5}{2} : u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{2} = \frac{1 - 5}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

解答は

u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{1 - 5}{2}, sin (x) = - \frac{1 - 5}{2}

sin (x) = 0, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{1 - 5}{2}, sin (x) = - \frac{1 - 5}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{1 + 5}{2} :

解なし

sin (x) = \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = - \frac{1 + 5}{2} :

解なし

sin (x) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = \frac{1 - 5}{2} : x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = \frac{1 - 5}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1 - 5}{2}

以下の一般解

sin (x) = \frac{1 - 5}{2}

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

x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2} : x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{1 - 5}{2}

以下の一般解

sin (x) = - \frac{1 - 5}{2}

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

x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn, x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

equationは以下で未定義のため:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

x = 2 πn, x = π + 2 πn

グラフ

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