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人気のある三角関数 >

3cot(x)=2sin(x)

解

3 cot (x) = 2 sin (x)

解

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

+1

度

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

解答ステップ

3 cot (x) = 2 sin (x)

両辺から

2 sin (x)

を引く

3 cot (x) - 2 sin (x) = 0

サイン, コサインで表わす

3 \cdot \frac{cos ( x )}{sin ( x )} - 2 sin (x) = 0

簡素化

3 \cdot \frac{cos ( x )}{sin ( x )} - 2 sin (x) : \frac{3 cos ( x ) - 2 sin ^{2} ( x )}{sin ( x )}

3 \cdot \frac{cos ( x )}{sin ( x )} - 2 sin (x)

乗じる

3 \cdot \frac{cos ( x )}{sin ( x )} : \frac{3 cos ( x )}{sin ( x )}

3 \cdot \frac{cos ( x )}{sin ( x )}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( x ) \cdot 3}{sin ( x )}

= \frac{3 cos ( x )}{sin ( x )} - 2 sin (x)

元を分数に変換する:

2 sin (x) = \frac{2 sin ( x ) sin ( x )}{sin ( x )}

= \frac{cos ( x ) \cdot 3}{sin ( x )} - \frac{2 sin ( x ) sin ( x )}{sin ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ( x ) \cdot 3 - 2 sin ( x ) sin ( x )}{sin ( x )}

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

cos (x) \cdot 3 - 2 sin (x) sin (x)

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

2 sin (x) sin (x)

指数の規則を適用する:

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

sin (x) sin (x) = sin^{1 + 1} (x)

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

数を足す:

1 + 1 = 2

= 2 sin^{2} (x)

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

= \frac{3 cos ( x ) - 2 sin ^{2} ( x )}{sin ( x )}

\frac{3 cos ( x ) - 2 sin ^{2} ( x )}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

両辺に

2 sin^{2} (x)

を足す

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

両辺を2乗する

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

両辺から

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

を引く

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

因数

9 cos^{2} (x) - 4 sin^{4} (x) : (3 cos (x) + 2 sin^{2} (x)) (3 cos (x) - 2 sin^{2} (x))

9 cos^{2} (x) - 4 sin^{4} (x)

9 cos^{2} (x) - 4 sin^{4} (x)

を書き換え

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

9 cos^{2} (x) - 4 sin^{4} (x)

9

を書き換え

3^{2}

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

4

を書き換え

2^{2}

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

指数の規則を適用する:

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

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

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

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

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

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

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

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

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

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

2乗の差の公式を適用する:

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

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

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

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

各部分を別個に解く

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

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

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

三角関数の公式を使用して書き換える

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

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

(1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

置換で解く

(1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

仮定:

cos (x) = u

(1 - u^{2}) \cdot 2 + 3 u = 0

(1 - u^{2}) \cdot 2 + 3 u = 0 : u = - \frac{1}{2}, u = 2

(1 - u^{2}) \cdot 2 + 3 u = 0

拡張

(1 - u^{2}) \cdot 2 + 3 u : 2 - 2 u^{2} + 3 u

(1 - u^{2}) \cdot 2 + 3 u

= 2 (1 - u^{2}) + 3 u

拡張

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数を乗じる:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= 2 - 2 u^{2} + 3 u

2 - 2 u^{2} + 3 u = 0

標準的な形式で書く

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

- 2 u^{2} + 3 u + 2 = 0

解くとthe二次式

- 2 u^{2} + 3 u + 2 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 3, c = 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

3^{2} - 4 (- 2) \cdot 2 = 5

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

規則を適用

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数を足す:

9 + 16 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 3 + 5}{2 ( - 2 )}, u_{2} = \frac{- 3 - 5}{2 ( - 2 )}

u = \frac{- 3 + 5}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 3 + 5}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 + 5}{- 2 \cdot 2}

数を足す/引く:

- 3 + 5 = 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{- 3 - 5}{2 ( - 2 )} : 2

\frac{- 3 - 5}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 - 5}{- 2 \cdot 2}

数を引く:

- 3 - 5 = - 8

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 8}{- 4}

分数の規則を適用する:

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

= \frac{8}{4}

数を割る:

\frac{8}{4} = 2

= 2

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

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) = 2 :

解なし

cos (x) = 2

- 1 \leq cos (x) \leq 1

解なし

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

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

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

三角関数の公式を使用して書き換える

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

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

- (1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

置換で解く

- (1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

仮定:

cos (x) = u

- (1 - u^{2}) \cdot 2 + 3 u = 0

- (1 - u^{2}) \cdot 2 + 3 u = 0 : u = \frac{1}{2}, u = - 2

- (1 - u^{2}) \cdot 2 + 3 u = 0

拡張

- (1 - u^{2}) \cdot 2 + 3 u : - 2 + 2 u^{2} + 3 u

- (1 - u^{2}) \cdot 2 + 3 u

= - 2 (1 - u^{2}) + 3 u

拡張

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

= - 2 \cdot 1 - (- 2) u^{2}

マイナス・プラスの規則を適用する

- (- a) = a

= - 2 \cdot 1 + 2 u^{2}

数を乗じる:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

= - 2 + 2 u^{2} + 3 u

- 2 + 2 u^{2} + 3 u = 0

標準的な形式で書く

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

2 u^{2} + 3 u - 2 = 0

解くとthe二次式

2 u^{2} + 3 u - 2 = 0

二次Equationの公式:

次の場合:

a = 2, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 2) = 5

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

規則を適用

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数を足す:

9 + 16 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

数を足す/引く:

- 3 + 5 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

数を引く:

- 3 - 5 = - 8

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 8}{4}

分数の規則を適用する:

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

= - \frac{8}{4}

数を割る:

\frac{8}{4} = 2

= - 2

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

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

cos (x) = - 2 :

解なし

cos (x) = - 2

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

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 = \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

3 cot (x) = 2 sin (x)

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

3 cot (x) = 2 sin (x)

の挿入向け

x = \frac{2 π}{3} + 2 π 1

3 cot (\frac{2 π}{3} + 2 π 1) = 2 sin (\frac{2 π}{3} + 2 π 1)

改良

- 1.73205 \dots = 1.73205 \dots

\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

3 cot (x) = 2 sin (x)

の挿入向け

x = \frac{4 π}{3} + 2 π 1

3 cot (\frac{4 π}{3} + 2 π 1) = 2 sin (\frac{4 π}{3} + 2 π 1)

改良

1.73205 \dots = - 1.73205 \dots

\Rightarrow 偽

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

3 cot (x) = 2 sin (x)

の挿入向け

x = \frac{π}{3} + 2 π 1

3 cot (\frac{π}{3} + 2 π 1) = 2 sin (\frac{π}{3} + 2 π 1)

改良

1.73205 \dots = 1.73205 \dots

\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

3 cot (x) = 2 sin (x)

の挿入向け

x = \frac{5 π}{3} + 2 π 1

3 cot (\frac{5 π}{3} + 2 π 1) = 2 sin (\frac{5 π}{3} + 2 π 1)

改良

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

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

グラフ

人気の例

tan (a) = \frac{7}{12}

sin(5x)=cos(3x)

sin (5 x) = cos (3 x)

sin^{3} (x) = \frac{1}{8}

solvefor y,sin(y^2)=x

so l v e f or y, sin (y^{2}) = x

cos(7x)=-(sqrt(3))/2

cos (7 x) = - \frac{3}{2}