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人気のある >

3tan(x)=2cos(x)

解

3 tan (x) = 2 cos (x)

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

解答ステップ

3 tan (x) = 2 cos (x)

両辺から

2 cos (x)

を引く

3 tan (x) - 2 cos (x) = 0

サイン, コサインで表わす

- 2 cos (x) + 3 tan (x)

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

簡素化

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

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

乗じる

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

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

分数を乗じる:

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

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

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

元を分数に変換する:

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

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

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

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

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

2 cos (x) cos (x)

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

置換で解く

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

仮定:

sin (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:

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 = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

以下の一般解

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

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - 2 :

解なし

sin (x) = - 2

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

人気の例

6tan^3(x)=2tan(x)

6 tan^{3} (x) = 2 tan (x)

tan^2(θ)=tan(θ)

tan^{2} (θ) = tan (θ)

sin(2θ)=-sin(θ)

sin (2 θ) = - sin (θ)

sin (x) = 2

cos^{2} (θ) - 1 = 0