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人気のある三角関数 >

2cos^2(x)=11cos(x)+1

解

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

解

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

+1

度

x = 95.13220 \dots^{\circ} + 36 0^{\circ} n, x = - 95.13220 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

2 u^{2} = 11 u + 1

2 u^{2} = 11 u + 1 : u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

2 u^{2} = 11 u + 1

1

を左側に移動します

2 u^{2} = 11 u + 1

両辺から

1

を引く

2 u^{2} - 1 = 11 u + 1 - 1

簡素化

2 u^{2} - 1 = 11 u

2 u^{2} - 1 = 11 u

11 u

を左側に移動します

2 u^{2} - 1 = 11 u

両辺から

11 u

を引く

2 u^{2} - 1 - 11 u = 11 u - 11 u

簡素化

2 u^{2} - 1 - 11 u = 0

2 u^{2} - 1 - 11 u = 0

標準的な形式で書く

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

2 u^{2} - 11 u - 1 = 0

解くとthe二次式

2 u^{2} - 11 u - 1 = 0

二次Equationの公式:

次の場合:

a = 2, b = - 11, c = - 1

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 11)^{2} - 4 \cdot 2 (- 1) = 129

(- 11)^{2} - 4 \cdot 2 (- 1)

規則を適用

- (- a) = a

= (- 11)^{2} + 4 \cdot 2 \cdot 1

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 11)^{2} = 1 1^{2}

= 1 1^{2} + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 1^{2} + 8

1 1^{2} = 121

= 121 + 8

数を足す:

121 + 8 = 129

= 129

u_{1, 2} = \frac{- ( - 11 ) \pm 129}{2 \cdot 2}

解を分離する

u_{1} = \frac{- ( - 11 ) + 129}{2 \cdot 2}, u_{2} = \frac{- ( - 11 ) - 129}{2 \cdot 2}

u = \frac{- ( - 11 ) + 129}{2 \cdot 2} : \frac{11 + 129}{4}

\frac{- ( - 11 ) + 129}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{11 + 129}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{11 + 129}{4}

u = \frac{- ( - 11 ) - 129}{2 \cdot 2} : \frac{11 - 129}{4}

\frac{- ( - 11 ) - 129}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{11 - 129}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{11 - 129}{4}

二次equationの解:

u = \frac{11 + 129}{4}, u = \frac{11 - 129}{4}

代用を戻す

u = cos (x)

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4}, cos (x) = \frac{11 - 129}{4}

cos (x) = \frac{11 + 129}{4} :

解なし

cos (x) = \frac{11 + 129}{4}

- 1 \leq cos (x) \leq 1

解なし

cos (x) = \frac{11 - 129}{4} : x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

cos (x) = \frac{11 - 129}{4}

三角関数の逆数プロパティを適用する

cos (x) = \frac{11 - 129}{4}

以下の一般解

cos (x) = \frac{11 - 129}{4}

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

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{11 - 129}{4}) + 2 πn, x = - arccos (\frac{11 - 129}{4}) + 2 πn

10進法形式で解を証明する

x = 1.66037 \dots + 2 πn, x = - 1.66037 \dots + 2 πn

グラフ

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