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人気のある >

8cos^2(x)-2cos(x)=1

解

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

解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

+1

度

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\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

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

置換で解く

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

仮定:

cos (x) = u

8 u^{2} - 2 u = 1

8 u^{2} - 2 u = 1 : u = \frac{1}{2}, u = - \frac{1}{4}

8 u^{2} - 2 u = 1

1

を左側に移動します

8 u^{2} - 2 u = 1

両辺から

1

を引く

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

簡素化

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 8, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 8 (- 1) = 6

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

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

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

= 2^{2} + 4 \cdot 8 \cdot 1

数を乗じる:

4 \cdot 8 \cdot 1 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 8}

解を分離する

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 8}

u = \frac{- ( - 2 ) + 6}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 2 ) + 6}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{2 + 6}{2 \cdot 8}

数を足す:

2 + 6 = 8

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

数を乗じる:

2 \cdot 8 = 16

= \frac{8}{16}

共通因数を約分する:

8

= \frac{1}{2}

u = \frac{- ( - 2 ) - 6}{2 \cdot 8} : - \frac{1}{4}

\frac{- ( - 2 ) - 6}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{2 - 6}{2 \cdot 8}

数を引く:

2 - 6 = - 4

= \frac{- 4}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{- 4}{16}

分数の規則を適用する:

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

= - \frac{4}{16}

共通因数を約分する:

4

= - \frac{1}{4}

二次equationの解:

u = \frac{1}{2}, u = - \frac{1}{4}

代用を戻す

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{4}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{4}

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

以下の一般解

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

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

人気の例

tan (x) = 1

tan(x)+sec(x)=1

tan (x) + sec (x) = 1

arctan (\frac{3}{4})

cos^2(x)+2cos(x)=3

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

tan (10 5^{\circ})