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人気のある三角関数 >

6(cos(b))^2+cos(b)-1=0

解

6 (cos (b))^{2} + cos (b) - 1 = 0

解

b = 1.23095 \dots + 2 πn, b = 2 π - 1.23095 \dots + 2 πn, b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

+1

度

b = 70.52877 \dots^{\circ} + 36 0^{\circ} n, b = 289.47122 \dots^{\circ} + 36 0^{\circ} n, b = 12 0^{\circ} + 36 0^{\circ} n, b = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

6 (cos (b))^{2} + cos (b) - 1 = 0

置換で解く

6 (cos (b))^{2} + cos (b) - 1 = 0

仮定:

cos (b) = u

6 u^{2} + u - 1 = 0

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

6 u^{2} + u - 1 = 0

解くとthe二次式

6 u^{2} + u - 1 = 0

二次Equationの公式:

次の場合:

a = 6, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 1) = 5

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

数を乗じる:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

数を足す:

1 + 24 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 6}

解を分離する

u_{1} = \frac{- 1 + 5}{2 \cdot 6}, u_{2} = \frac{- 1 - 5}{2 \cdot 6}

u = \frac{- 1 + 5}{2 \cdot 6} : \frac{1}{3}

\frac{- 1 + 5}{2 \cdot 6}

数を足す/引く:

- 1 + 5 = 4

= \frac{4}{2 \cdot 6}

数を乗じる:

2 \cdot 6 = 12

= \frac{4}{12}

共通因数を約分する:

4

= \frac{1}{3}

u = \frac{- 1 - 5}{2 \cdot 6} : - \frac{1}{2}

\frac{- 1 - 5}{2 \cdot 6}

数を引く:

- 1 - 5 = - 6

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

数を乗じる:

2 \cdot 6 = 12

= \frac{- 6}{12}

分数の規則を適用する:

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

= - \frac{6}{12}

共通因数を約分する:

6

= - \frac{1}{2}

二次equationの解:

u = \frac{1}{3}, u = - \frac{1}{2}

代用を戻す

u = cos (b)

cos (b) = \frac{1}{3}, cos (b) = - \frac{1}{2}

cos (b) = \frac{1}{3}, cos (b) = - \frac{1}{2}

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

cos (b) = - \frac{1}{2} : b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

cos (b) = - \frac{1}{2}

以下の一般解

cos (b) = - \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}

b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

すべての解を組み合わせる

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

10進法形式で解を証明する

b = 1.23095 \dots + 2 πn, b = 2 π - 1.23095 \dots + 2 πn, b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

グラフ

人気の例

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sin (θ) = \frac{1.33}{2.42}

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