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人気のある三角関数 >

6cos^2(x)+5cos(x)+1=0

解

6 cos^{2} (x) + 5 cos (x) + 1 = 0

解

x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度

x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

6 cos^{2} (x) + 5 cos (x) + 1 = 0

置換で解く

6 cos^{2} (x) + 5 cos (x) + 1 = 0

仮定:

cos (x) = u

6 u^{2} + 5 u + 1 = 0

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

6 u^{2} + 5 u + 1 = 0

解くとthe二次式

6 u^{2} + 5 u + 1 = 0

二次Equationの公式:

次の場合:

a = 6, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

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

5^{2} - 4 \cdot 6 \cdot 1

数を乗じる:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数を引く:

25 - 24 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

数を足す/引く:

- 5 + 1 = - 4

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

数を乗じる:

2 \cdot 6 = 12

= \frac{- 4}{12}

分数の規則を適用する:

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

= - \frac{4}{12}

共通因数を約分する:

4

= - \frac{1}{3}

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

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

数を引く:

- 5 - 1 = - 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 (x)

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

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

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

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

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

グラフ

人気の例

tan^4(x)+1=4tan^2(x)-2

tan^{4} (x) + 1 = 4 tan^{2} (x) - 2

tan^2(x)=sec(x)

tan^{2} (x) = sec (x)

solvefor x,sin(x)-sin(2x)=0

so l v e f or x, sin (x) - sin (2 x) = 0

sin(θ)= 4/5 ,cos(θ)

sin (θ) = \frac{4}{5}, cos (θ)

sin(6x)=cos(3x)

sin (6 x) = cos (3 x)