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人気のある三角関数 >

12cos^2(x)-cos(x)-1=0

解

12 cos^{2} (x) - cos (x) - 1 = 0

解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

+1

度

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

12 cos^{2} (x) - cos (x) - 1 = 0

置換で解く

12 cos^{2} (x) - cos (x) - 1 = 0

仮定:

cos (x) = u

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

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 12, b = - 1, c = - 1

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

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

(- 1)^{2} - 4 \cdot 12 (- 1) = 7

(- 1)^{2} - 4 \cdot 12 (- 1)

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 12 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 12 \cdot 1 = 48

4 \cdot 12 \cdot 1

数を乗じる:

4 \cdot 12 \cdot 1 = 48

= 48

= 1 + 48

数を足す:

1 + 48 = 49

= 49

数を因数に分解する:

49 = 7^{2}

= 7^{2}

累乗根の規則を適用する:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 12}

解を分離する

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 12}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 12}

u = \frac{- ( - 1 ) + 7}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 12}

規則を適用

- (- a) = a

= \frac{1 + 7}{2 \cdot 12}

数を足す:

1 + 7 = 8

= \frac{8}{2 \cdot 12}

数を乗じる:

2 \cdot 12 = 24

= \frac{8}{24}

共通因数を約分する:

8

= \frac{1}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 12} : - \frac{1}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 12}

規則を適用

- (- a) = a

= \frac{1 - 7}{2 \cdot 12}

数を引く:

1 - 7 = - 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}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{3}

以下の一般解

cos (x) = \frac{1}{3}

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

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

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

すべての解を組み合わせる

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

10進法形式で解を証明する

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

グラフ

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