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人気のある三角関数 >

4cos(2x)=1-3cos(x)

解

4 cos (2 x) = 1 - 3 cos (x)

解

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

+1

度

x = 51.31781 \dots^{\circ} + 36 0^{\circ} n, x = 308.68218 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

4 cos (2 x) = 1 - 3 cos (x)

両辺から

1 - 3 cos (x)

を引く

4 cos (2 x) - 1 + 3 cos (x) = 0

三角関数の公式を使用して書き換える

- 1 + 3 cos (x) + 4 cos (2 x)

2倍角の公式を使用:

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

= - 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1)

簡素化

- 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) + 3 cos (x) - 5

- 1 + 3 cos (x) + 4 (2 cos^{2} (x) - 1)

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

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

簡素化

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

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

数を乗じる:

4 \cdot 2 = 8

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

数を乗じる:

4 \cdot 1 = 4

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

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

= - 1 + 3 cos (x) + 8 cos^{2} (x) - 4

簡素化

- 1 + 3 cos (x) + 8 cos^{2} (x) - 4 : 8 cos^{2} (x) + 3 cos (x) - 5

- 1 + 3 cos (x) + 8 cos^{2} (x) - 4

条件のようなグループ

= 3 cos (x) + 8 cos^{2} (x) - 1 - 4

数を引く:

- 1 - 4 = - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

= 8 cos^{2} (x) + 3 cos (x) - 5

- 5 + 3 cos (x) + 8 cos^{2} (x) = 0

置換で解く

- 5 + 3 cos (x) + 8 cos^{2} (x) = 0

仮定:

cos (x) = u

- 5 + 3 u + 8 u^{2} = 0

- 5 + 3 u + 8 u^{2} = 0 : u = \frac{5}{8}, u = - 1

- 5 + 3 u + 8 u^{2} = 0

標準的な形式で書く

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

8 u^{2} + 3 u - 5 = 0

解くとthe二次式

8 u^{2} + 3 u - 5 = 0

二次Equationの公式:

次の場合:

a = 8, b = 3, c = - 5

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

3^{2} - 4 \cdot 8 (- 5) = 13

3^{2} - 4 \cdot 8 (- 5)

規則を適用

- (- a) = a

= 3^{2} + 4 \cdot 8 \cdot 5

数を乗じる:

4 \cdot 8 \cdot 5 = 160

= 3^{2} + 160

3^{2} = 9

= 9 + 160

数を足す:

9 + 160 = 169

= 169

数を因数に分解する:

169 = 1 3^{2}

= 1 3^{2}

累乗根の規則を適用する:

n a^{n} = a

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 8}

解を分離する

u_{1} = \frac{- 3 + 13}{2 \cdot 8}, u_{2} = \frac{- 3 - 13}{2 \cdot 8}

u = \frac{- 3 + 13}{2 \cdot 8} : \frac{5}{8}

\frac{- 3 + 13}{2 \cdot 8}

数を足す/引く:

- 3 + 13 = 10

= \frac{10}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{10}{16}

共通因数を約分する:

2

= \frac{5}{8}

u = \frac{- 3 - 13}{2 \cdot 8} : - 1

\frac{- 3 - 13}{2 \cdot 8}

数を引く:

- 3 - 13 = - 16

= \frac{- 16}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{- 16}{16}

分数の規則を適用する:

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

= - \frac{16}{16}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

u = \frac{5}{8}, u = - 1

代用を戻す

u = cos (x)

cos (x) = \frac{5}{8}, cos (x) = - 1

cos (x) = \frac{5}{8}, cos (x) = - 1

cos (x) = \frac{5}{8} : x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

cos (x) = \frac{5}{8}

三角関数の逆数プロパティを適用する

cos (x) = \frac{5}{8}

以下の一般解

cos (x) = \frac{5}{8}

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

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn

x = arccos (\frac{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 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{5}{8}) + 2 πn, x = 2 π - arccos (\frac{5}{8}) + 2 πn, x = π + 2 πn

10進法形式で解を証明する

x = 0.89566 \dots + 2 πn, x = 2 π - 0.89566 \dots + 2 πn, x = π + 2 πn

グラフ

人気の例

0=32pisec(x^2)tan(x)

0 = 32 π sec (x^{2}) tan (x)

1*sin(54.86)=2.451*sin(x)

1 \cdot sin (54.8 6^{\circ}) = 2.451 \cdot sin (x)

(cot(θ))/(csc(θ))=sin(θ)

\frac{cot ( θ )}{csc ( θ )} = sin (θ)

cos (t) = - \frac{2}{3}

0 = 4 cos (4 x)