ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

cos^4(x)=1-8sin^2(x)cos^2(x)

解

cos^{4} (x) = 1 - 8 sin^{2} (x) cos^{2} (x)

解

x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn, x = 2 πn, x = π + 2 πn

+1

度

x = 67.79234 \dots^{\circ} + 36 0^{\circ} n, x = 292.20765 \dots^{\circ} + 36 0^{\circ} n, x = 112.20765 \dots^{\circ} + 36 0^{\circ} n, x = - 112.20765 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

cos^{4} (x) = 1 - 8 sin^{2} (x) cos^{2} (x)

両辺から

1 - 8 sin^{2} (x) cos^{2} (x)

を引く

cos^{4} (x) - 1 + 8 sin^{2} (x) cos^{2} (x) = 0

三角関数の公式を使用して書き換える

- 1 + cos^{4} (x) + 8 cos^{2} (x) sin^{2} (x)

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)

= - 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x))

簡素化

- 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x)) : - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x))

拡張

8 cos^{2} (x) (1 - cos^{2} (x)) : 8 cos^{2} (x) - 8 cos^{4} (x)

8 cos^{2} (x) (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 8 cos^{2} (x), b = 1, c = cos^{2} (x)

= 8 cos^{2} (x) \cdot 1 - 8 cos^{2} (x) cos^{2} (x)

= 8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x)

簡素化

8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x) : 8 cos^{2} (x) - 8 cos^{4} (x)

8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x)

8 \cdot 1 \cdot cos^{2} (x) = 8 cos^{2} (x)

8 \cdot 1 \cdot cos^{2} (x)

数を乗じる:

8 \cdot 1 = 8

= 8 cos^{2} (x)

8 cos^{2} (x) cos^{2} (x) = 8 cos^{4} (x)

8 cos^{2} (x) cos^{2} (x)

指数の規則を適用する:

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

cos^{2} (x) cos^{2} (x) = cos^{2 + 2} (x)

= 8 cos^{2 + 2} (x)

数を足す:

2 + 2 = 4

= 8 cos^{4} (x)

= 8 cos^{2} (x) - 8 cos^{4} (x)

= 8 cos^{2} (x) - 8 cos^{4} (x)

= - 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x)

簡素化

- 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x) : - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x)

条件のようなグループ

= cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x) - 1

類似した元を足す:

cos^{4} (x) - 8 cos^{4} (x) = - 7 cos^{4} (x)

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 - 7 cos^{4} (x) + 8 cos^{2} (x) = 0

置換で解く

- 1 - 7 cos^{4} (x) + 8 cos^{2} (x) = 0

仮定:

cos (x) = u

- 1 - 7 u^{4} + 8 u^{2} = 0

- 1 - 7 u^{4} + 8 u^{2} = 0 : u = \frac{1}{7}, u = - \frac{1}{7}, u = 1, u = - 1

- 1 - 7 u^{4} + 8 u^{2} = 0

標準的な形式で書く

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

- 7 u^{4} + 8 u^{2} - 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

- 7 v^{2} + 8 v - 1 = 0

解く

- 7 v^{2} + 8 v - 1 = 0 : v = \frac{1}{7}, v = 1

- 7 v^{2} + 8 v - 1 = 0

解くとthe二次式

- 7 v^{2} + 8 v - 1 = 0

二次Equationの公式:

次の場合:

a = - 7, b = 8, c = - 1

v_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 7 ) ( - 1 )}{2 ( - 7 )}

v_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 7 ) ( - 1 )}{2 ( - 7 )}

8^{2} - 4 (- 7) (- 1) = 6

8^{2} - 4 (- 7) (- 1)

規則を適用

- (- a) = a

= 8^{2} - 4 \cdot 7 \cdot 1

数を乗じる:

4 \cdot 7 \cdot 1 = 28

= 8^{2} - 28

8^{2} = 64

= 64 - 28

数を引く:

64 - 28 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

v_{1, 2} = \frac{- 8 \pm 6}{2 ( - 7 )}

解を分離する

v_{1} = \frac{- 8 + 6}{2 ( - 7 )}, v_{2} = \frac{- 8 - 6}{2 ( - 7 )}

v = \frac{- 8 + 6}{2 ( - 7 )} : \frac{1}{7}

\frac{- 8 + 6}{2 ( - 7 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 + 6}{- 2 \cdot 7}

数を足す/引く:

- 8 + 6 = - 2

= \frac{- 2}{- 2 \cdot 7}

数を乗じる:

2 \cdot 7 = 14

= \frac{- 2}{- 14}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{14}

共通因数を約分する:

2

= \frac{1}{7}

v = \frac{- 8 - 6}{2 ( - 7 )} : 1

\frac{- 8 - 6}{2 ( - 7 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 - 6}{- 2 \cdot 7}

数を引く:

- 8 - 6 = - 14

= \frac{- 14}{- 2 \cdot 7}

数を乗じる:

2 \cdot 7 = 14

= \frac{- 14}{- 14}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

= \frac{14}{14}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

v = \frac{1}{7}, v = 1

v = \frac{1}{7}, v = 1

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{1}{7} : u = \frac{1}{7}, u = - \frac{1}{7}

u^{2} = \frac{1}{7}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{7}, u = - \frac{1}{7}

解く

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

解答は

u = \frac{1}{7}, u = - \frac{1}{7}, u = 1, u = - 1

代用を戻す

u = cos (x)

cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}, cos (x) = 1, cos (x) = - 1

cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}, cos (x) = 1, cos (x) = - 1

cos (x) = \frac{1}{7} : x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = \frac{1}{7}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{7}

以下の一般解

cos (x) = \frac{1}{7}

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

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7} : x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1}{7}

以下の一般解

cos (x) = - \frac{1}{7}

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

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

以下の一般解

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn, x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn, x = 2 πn, x = π + 2 πn

10進法形式で解を証明する

x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn, x = 2 πn, x = π + 2 πn

グラフ

人気の例

tan (x) = \frac{10}{18}

((1+cot^2(x)))/(cos^2(x))=cot^2(x)

\frac{( 1 + cot ^{2} ( x ) )}{cos ^{2} ( x )} = cot^{2} (x)

sin (2 p + 1) = - 1

solvefor y,a*z=5sin(2y)

so l v e f or y, a \cdot z = 5 sin (2 y)

sin^3(x)=sin^2(x)

sin^{3} (x) = sin^{2} (x)