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人気のある三角関数 >

-6cos(2x)+23cos(x)-11=0

解

- 6 cos (2 x) + 23 cos (x) - 11 = 0

解

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

+1

度

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

- 6 cos (2 x) + 23 cos (x) - 11 = 0

三角関数の公式を使用して書き換える

- 11 + 23 cos (x) - 6 cos (2 x)

2倍角の公式を使用:

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

= - 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1)

簡素化

- 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1) : 23 cos (x) - 12 cos^{2} (x) - 5

- 11 + 23 cos (x) - 6 (2 cos^{2} (x) - 1)

拡張

- 6 (2 cos^{2} (x) - 1) : - 12 cos^{2} (x) + 6

- 6 (2 cos^{2} (x) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = - 6, b = 2 cos^{2} (x), c = 1

= - 6 \cdot 2 cos^{2} (x) - (- 6) \cdot 1

マイナス・プラスの規則を適用する

- (- a) = a

= - 6 \cdot 2 cos^{2} (x) + 6 \cdot 1

簡素化

- 6 \cdot 2 cos^{2} (x) + 6 \cdot 1 : - 12 cos^{2} (x) + 6

- 6 \cdot 2 cos^{2} (x) + 6 \cdot 1

数を乗じる:

6 \cdot 2 = 12

= - 12 cos^{2} (x) + 6 \cdot 1

数を乗じる:

6 \cdot 1 = 6

= - 12 cos^{2} (x) + 6

= - 12 cos^{2} (x) + 6

= - 11 + 23 cos (x) - 12 cos^{2} (x) + 6

簡素化

- 11 + 23 cos (x) - 12 cos^{2} (x) + 6 : 23 cos (x) - 12 cos^{2} (x) - 5

- 11 + 23 cos (x) - 12 cos^{2} (x) + 6

条件のようなグループ

= 23 cos (x) - 12 cos^{2} (x) - 11 + 6

数を足す/引く:

- 11 + 6 = - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

= 23 cos (x) - 12 cos^{2} (x) - 5

- 5 - 12 cos^{2} (x) + 23 cos (x) = 0

置換で解く

- 5 - 12 cos^{2} (x) + 23 cos (x) = 0

仮定:

cos (x) = u

- 5 - 12 u^{2} + 23 u = 0

- 5 - 12 u^{2} + 23 u = 0 : u = \frac{1}{4}, u = \frac{5}{3}

- 5 - 12 u^{2} + 23 u = 0

標準的な形式で書く

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

- 12 u^{2} + 23 u - 5 = 0

解くとthe二次式

- 12 u^{2} + 23 u - 5 = 0

二次Equationの公式:

次の場合:

a = - 12, b = 23, c = - 5

u_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 ( - 12 ) ( - 5 )}{2 ( - 12 )}

u_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 ( - 12 ) ( - 5 )}{2 ( - 12 )}

2 3^{2} - 4 (- 12) (- 5) = 17

2 3^{2} - 4 (- 12) (- 5)

規則を適用

- (- a) = a

= 2 3^{2} - 4 \cdot 12 \cdot 5

数を乗じる:

4 \cdot 12 \cdot 5 = 240

= 2 3^{2} - 240

2 3^{2} = 529

= 529 - 240

数を引く:

529 - 240 = 289

= 289

数を因数に分解する:

289 = 1 7^{2}

= 1 7^{2}

累乗根の規則を適用する:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 23 \pm 17}{2 ( - 12 )}

解を分離する

u_{1} = \frac{- 23 + 17}{2 ( - 12 )}, u_{2} = \frac{- 23 - 17}{2 ( - 12 )}

u = \frac{- 23 + 17}{2 ( - 12 )} : \frac{1}{4}

\frac{- 23 + 17}{2 ( - 12 )}

括弧を削除する:

(- a) = - a

= \frac{- 23 + 17}{- 2 \cdot 12}

数を足す/引く:

- 23 + 17 = - 6

= \frac{- 6}{- 2 \cdot 12}

数を乗じる:

2 \cdot 12 = 24

= \frac{- 6}{- 24}

分数の規則を適用する:

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

= \frac{6}{24}

共通因数を約分する:

6

= \frac{1}{4}

u = \frac{- 23 - 17}{2 ( - 12 )} : \frac{5}{3}

\frac{- 23 - 17}{2 ( - 12 )}

括弧を削除する:

(- a) = - a

= \frac{- 23 - 17}{- 2 \cdot 12}

数を引く:

- 23 - 17 = - 40

= \frac{- 40}{- 2 \cdot 12}

数を乗じる:

2 \cdot 12 = 24

= \frac{- 40}{- 24}

分数の規則を適用する:

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

= \frac{40}{24}

共通因数を約分する:

8

= \frac{5}{3}

二次equationの解:

u = \frac{1}{4}, u = \frac{5}{3}

代用を戻す

u = cos (x)

cos (x) = \frac{1}{4}, cos (x) = \frac{5}{3}

cos (x) = \frac{1}{4}, cos (x) = \frac{5}{3}

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

cos (x) = \frac{1}{4}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{4}

以下の一般解

cos (x) = \frac{1}{4}

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

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

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

cos (x) = \frac{5}{3} :

解なし

cos (x) = \frac{5}{3}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

グラフ

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