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 >

2(sin(x))^2-cos(x)-1=0

解

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

解

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度

x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺に

cos (x)

を足す

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

両辺を2乗する

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

両辺から

cos^{2} (x)

を引く

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

因数

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

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

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

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

改良

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

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

各部分を別個に解く

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

2 sin^{2} (x) - 1 + cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

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

三角関数の公式を使用して書き換える

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

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

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

条件のようなグループ

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

数を足す/引く:

- 1 + 2 = 1

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = 1, c = 1

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

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

数を足す/引く:

- 1 + 3 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{- 4}

分数の規則を適用する:

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

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

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

数を引く:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

分数の規則を適用する:

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

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

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

代用を戻す

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = 1

cos (x) = - \frac{1}{2}, cos (x) = 1

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

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

以下の一般解

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 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

すべての解を組み合わせる

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn

2 sin^{2} (x) - 1 - cos (x) = 0 : x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

三角関数の公式を使用して書き換える

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

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

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

条件のようなグループ

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

数を足す/引く:

- 1 + 2 = 1

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

1 - u - 2 u^{2} = 0

1 - u - 2 u^{2} = 0 : u = - 1, u = \frac{1}{2}

1 - u - 2 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = - 1, c = 1

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

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

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

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

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \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 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a, - (- a) = a

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

数を足す:

1 + 3 = 4

= \frac{4}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{- 4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a, - (- a) = a

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

数を引く:

1 - 3 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

二次equationの解:

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

代用を戻す

u = cos (x)

cos (x) = - 1, cos (x) = \frac{1}{2}

cos (x) = - 1, cos (x) = \frac{1}{2}

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

以下の一般解

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

すべての解を組み合わせる

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

すべての解を組み合わせる

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2 πn, x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

元のequationに当てはめて解を検算する

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

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

\frac{2 π}{3} + 2 πn :

偽

\frac{2 π}{3} + 2 πn

挿入

n = 1

\frac{2 π}{3} + 2 π 1

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

の挿入向け

x = \frac{2 π}{3} + 2 π 1

2 (sin (\frac{2 π}{3} + 2 π 1))^{2} - cos (\frac{2 π}{3} + 2 π 1) - 1 = 0

改良

1 = 0

\Rightarrow 偽

解答を確認する

\frac{4 π}{3} + 2 πn :

偽

\frac{4 π}{3} + 2 πn

挿入

n = 1

\frac{4 π}{3} + 2 π 1

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

の挿入向け

x = \frac{4 π}{3} + 2 π 1

2 (sin (\frac{4 π}{3} + 2 π 1))^{2} - cos (\frac{4 π}{3} + 2 π 1) - 1 = 0

改良

1 = 0

\Rightarrow 偽

解答を確認する

2 πn :

偽

2 πn

挿入

n = 1

2 π 1

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

の挿入向け

x = 2 π 1

2 (sin (2 π 1))^{2} - cos (2 π 1) - 1 = 0

改良

- 2 = 0

\Rightarrow 偽

解答を確認する

π + 2 πn :

真

π + 2 πn

挿入

n = 1

π + 2 π 1

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

の挿入向け

x = π + 2 π 1

2 (sin (π + 2 π 1))^{2} - cos (π + 2 π 1) - 1 = 0

改良

0 = 0

\Rightarrow 真

解答を確認する

\frac{π}{3} + 2 πn :

真

\frac{π}{3} + 2 πn

挿入

n = 1

\frac{π}{3} + 2 π 1

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

の挿入向け

x = \frac{π}{3} + 2 π 1

2 (sin (\frac{π}{3} + 2 π 1))^{2} - cos (\frac{π}{3} + 2 π 1) - 1 = 0

改良

0 = 0

\Rightarrow 真

解答を確認する

\frac{5 π}{3} + 2 πn :

真

\frac{5 π}{3} + 2 πn

挿入

n = 1

\frac{5 π}{3} + 2 π 1

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

の挿入向け

x = \frac{5 π}{3} + 2 π 1

2 (sin (\frac{5 π}{3} + 2 π 1))^{2} - cos (\frac{5 π}{3} + 2 π 1) - 1 = 0

改良

0 = 0

\Rightarrow 真

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

グラフ

人気の例

arctan(3x)+arctan(x)= pi/4

arctan (3 x) + arctan (x) = \frac{π}{4}

4sin(θ)=sqrt(3)sec(θ),0<= θ<180

4 sin (θ) = 3 sec (θ), 0 \leq θ < 18 0^{\circ}

(sin(82))/(sin(x))=sqrt(5)

\frac{sin ( 8 2 ^{\circ} )}{sin ( x )} = 5

(2cos^2(x))/(2(1-sin(x))-cos^2(x))=0

\frac{2 cos ^{2} ( x )}{2 ( 1 - sin ( x ) ) - cos ^{2} ( x )} = 0

2 sin (x) = - 3