ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある >

sin^2(x)+cos^4(x)=n

解

sin^{2} (x) + cos^{4} (x) = n

解

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

解答ステップ

sin^{2} (x) + cos^{4} (x) = n

両辺から

n

を引く

sin^{2} (x) + cos^{4} (x) - n = 0

三角関数の公式を使用して書き換える

cos^{4} (x) + sin^{2} (x) - n

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^{4} (x) + 1 - cos^{2} (x) - n

1 - cos^{2} (x) + cos^{4} (x) - n = 0

置換で解く

1 - cos^{2} (x) + cos^{4} (x) - n = 0

仮定:

cos (x) = u

1 - u^{2} + u^{4} - n = 0

1 - u^{2} + u^{4} - n = 0 : u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}, u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

1 - u^{2} + u^{4} - n = 0

標準的な形式で書く

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

u^{4} - u^{2} + 1 - n = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} - v + 1 - n = 0

解く

v^{2} - v + 1 - n = 0 : v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

v^{2} - v + 1 - n = 0

解くとthe二次式

v^{2} - v + 1 - n = 0

二次Equationの公式:

次の場合:

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

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

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

簡素化

(- 1)^{2} - 4 \cdot 1 \cdot (1 - n) : 4 n - 3

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

(- 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 - n) = 4 (1 - n)

4 \cdot 1 \cdot (1 - n)

数を乗じる:

4 \cdot 1 = 4

= 4 (- n + 1)

= 1 - 4 (- n + 1)

拡張

1 - 4 (1 - n) : 4 n - 3

1 - 4 (1 - n)

拡張

- 4 (1 - n) : - 4 + 4 n

- 4 (1 - n)

分配法則を適用する:

a (b - c) = ab - a c

a = - 4, b = 1, c = n

= - 4 \cdot 1 - (- 4) n

マイナス・プラスの規則を適用する

- (- a) = a

= - 4 \cdot 1 + 4 n

数を乗じる:

4 \cdot 1 = 4

= - 4 + 4 n

= 1 - 4 + 4 n

数を引く:

1 - 4 = - 3

= 4 n - 3

= 4 n - 3

v_{1, 2} = \frac{- ( - 1 ) \pm 4 n - 3}{2 \cdot 1}

解を分離する

v_{1} = \frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1}

v = \frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1} : \frac{1 + 4 n - 3}{2}

\frac{- ( - 1 ) + 4 n - 3}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 + 4 n - 3}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 + 4 n - 3}{2}

v = \frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1} : \frac{1 - 4 n - 3}{2}

\frac{- ( - 1 ) - 4 n - 3}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 - 4 n - 3}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 - 4 n - 3}{2}

二次equationの解:

v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

v = \frac{1 + 4 n - 3}{2}, v = \frac{1 - 4 n - 3}{2}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{1 + 4 n - 3}{2} : u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}

u^{2} = \frac{1 + 4 n - 3}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}

解く

u^{2} = \frac{1 - 4 n - 3}{2} : u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

u^{2} = \frac{1 - 4 n - 3}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

解答は

u = \frac{1 + 4 n - 3}{2}, u = - \frac{1 + 4 n - 3}{2}, u = \frac{1 - 4 n - 3}{2}, u = - \frac{1 - 4 n - 3}{2}

代用を戻す

u = cos (x)

cos (x) = \frac{1 + 4 n - 3}{2}, cos (x) = - \frac{1 + 4 n - 3}{2}, cos (x) = \frac{1 - 4 n - 3}{2}, cos (x) = - \frac{1 - 4 n - 3}{2}

cos (x) = \frac{1 + 4 n - 3}{2}, cos (x) = - \frac{1 + 4 n - 3}{2}, cos (x) = \frac{1 - 4 n - 3}{2}, cos (x) = - \frac{1 - 4 n - 3}{2}

cos (x) = \frac{1 + 4 n - 3}{2} : x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 + 4 n - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1 + 4 n - 3}{2}

以下の一般解

cos (x) = \frac{1 + 4 n - 3}{2}

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

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 + 4 n - 3}{2} : x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 + 4 n - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1 + 4 n - 3}{2}

以下の一般解

cos (x) = - \frac{1 + 4 n - 3}{2}

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

x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 - 4 n - 3}{2} : x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = \frac{1 - 4 n - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1 - 4 n - 3}{2}

以下の一般解

cos (x) = \frac{1 - 4 n - 3}{2}

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

x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 - 4 n - 3}{2} : x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

cos (x) = - \frac{1 - 4 n - 3}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1 - 4 n - 3}{2}

以下の一般解

cos (x) = - \frac{1 - 4 n - 3}{2}

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

x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

すべての解を組み合わせる

x = arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 + 4 n - 3}{2} + 2 πk, x = arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos \frac{1 - 4 n - 3}{2} + 2 πk, x = arccos - \frac{1 - 4 n - 3}{2} + 2 πk, x = - arccos - \frac{1 - 4 n - 3}{2} + 2 πk

グラフ

人気の例

0 = sin (t)

csc^2(x)+sec^2(x)=1

csc^{2} (x) + sec^{2} (x) = 1

solvefor x,z=arctan(xy)解く

x, z = arctan (x y)

4(1+sin(θ))sin(θ)=3

4 (1 + sin (θ)) sin (θ) = 3

cos(x)+1=-cos(2x)

cos (x) + 1 = - cos (2 x)