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人気のある三角関数 >

10cos^2(x)+13cos(x)+4=0

解

10 cos^{2} (x) + 13 cos (x) + 4 = 0

解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

+1

度

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 143.13010 \dots^{\circ} + 36 0^{\circ} n, x = - 143.13010 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

10 cos^{2} (x) + 13 cos (x) + 4 = 0

置換で解く

10 cos^{2} (x) + 13 cos (x) + 4 = 0

仮定:

cos (x) = u

10 u^{2} + 13 u + 4 = 0

10 u^{2} + 13 u + 4 = 0 : u = - \frac{1}{2}, u = - \frac{4}{5}

10 u^{2} + 13 u + 4 = 0

解くとthe二次式

10 u^{2} + 13 u + 4 = 0

二次Equationの公式:

次の場合:

a = 10, b = 13, c = 4

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 10 \cdot 4}{2 \cdot 10}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 10 \cdot 4}{2 \cdot 10}

1 3^{2} - 4 \cdot 10 \cdot 4 = 3

1 3^{2} - 4 \cdot 10 \cdot 4

数を乗じる:

4 \cdot 10 \cdot 4 = 160

= 1 3^{2} - 160

1 3^{2} = 169

= 169 - 160

数を引く:

169 - 160 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 13 \pm 3}{2 \cdot 10}

解を分離する

u_{1} = \frac{- 13 + 3}{2 \cdot 10}, u_{2} = \frac{- 13 - 3}{2 \cdot 10}

u = \frac{- 13 + 3}{2 \cdot 10} : - \frac{1}{2}

\frac{- 13 + 3}{2 \cdot 10}

数を足す/引く:

- 13 + 3 = - 10

= \frac{- 10}{2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{- 10}{20}

分数の規則を適用する:

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

= - \frac{10}{20}

共通因数を約分する:

10

= - \frac{1}{2}

u = \frac{- 13 - 3}{2 \cdot 10} : - \frac{4}{5}

\frac{- 13 - 3}{2 \cdot 10}

数を引く:

- 13 - 3 = - 16

= \frac{- 16}{2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{- 16}{20}

分数の規則を適用する:

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

= - \frac{16}{20}

共通因数を約分する:

4

= - \frac{4}{5}

二次equationの解:

u = - \frac{1}{2}, u = - \frac{4}{5}

代用を戻す

u = cos (x)

cos (x) = - \frac{1}{2}, cos (x) = - \frac{4}{5}

cos (x) = - \frac{1}{2}, cos (x) = - \frac{4}{5}

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

cos (x) = - \frac{4}{5} : x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

cos (x) = - \frac{4}{5}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{4}{5}

以下の一般解

cos (x) = - \frac{4}{5}

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

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

すべての解を組み合わせる

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

10進法形式で解を証明する

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

グラフ

人気の例

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

csc^{2} (x) = \frac{4}{3}

cot^{2} (\frac{x}{2}) = 3

4cos(2x)+3cos(x)=1

4 cos (2 x) + 3 cos (x) = 1

0 = 2 sin (x)