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人気のある三角関数 >

10cos^2(x)+cos(x)=11sin^2(x)-9

解

10 cos^{2} (x) + cos (x) = 11 sin^{2} (x) - 9

解

x = 1.28104 \dots + 2 πn, x = 2 π - 1.28104 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

+1

度

x = 73.39845 \dots^{\circ} + 36 0^{\circ} n, x = 286.60154 \dots^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

10 cos^{2} (x) + cos (x) = 11 sin^{2} (x) - 9

両辺から

11 sin^{2} (x) - 9

を引く

10 cos^{2} (x) + cos (x) - 11 sin^{2} (x) + 9 = 0

三角関数の公式を使用して書き換える

9 + cos (x) + 10 cos^{2} (x) - 11 sin^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

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

= 9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x))

簡素化

9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x)) : 21 cos^{2} (x) + cos (x) - 2

9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x))

拡張

- 11 (1 - cos^{2} (x)) : - 11 + 11 cos^{2} (x)

- 11 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 11, b = 1, c = cos^{2} (x)

= - 11 \cdot 1 - (- 11) cos^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 11 \cdot 1 + 11 cos^{2} (x)

数を乗じる:

11 \cdot 1 = 11

= - 11 + 11 cos^{2} (x)

= 9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x)

簡素化

9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x) : 21 cos^{2} (x) + cos (x) - 2

9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x)

条件のようなグループ

= cos (x) + 10 cos^{2} (x) + 11 cos^{2} (x) + 9 - 11

類似した元を足す:

10 cos^{2} (x) + 11 cos^{2} (x) = 21 cos^{2} (x)

= cos (x) + 21 cos^{2} (x) + 9 - 11

数を足す/引く:

9 - 11 = - 2

= 21 cos^{2} (x) + cos (x) - 2

= 21 cos^{2} (x) + cos (x) - 2

= 21 cos^{2} (x) + cos (x) - 2

- 2 + cos (x) + 21 cos^{2} (x) = 0

置換で解く

- 2 + cos (x) + 21 cos^{2} (x) = 0

仮定:

cos (x) = u

- 2 + u + 21 u^{2} = 0

- 2 + u + 21 u^{2} = 0 : u = \frac{2}{7}, u = - \frac{1}{3}

- 2 + u + 21 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

21 u^{2} + u - 2 = 0

解くとthe二次式

21 u^{2} + u - 2 = 0

二次Equationの公式:

次の場合:

a = 21, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 21 ( - 2 )}{2 \cdot 21}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 21 ( - 2 )}{2 \cdot 21}

1^{2} - 4 \cdot 21 (- 2) = 13

1^{2} - 4 \cdot 21 (- 2)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 21 (- 2)

規則を適用

- (- a) = a

= 1 + 4 \cdot 21 \cdot 2

数を乗じる:

4 \cdot 21 \cdot 2 = 168

= 1 + 168

数を足す:

1 + 168 = 169

= 169

数を因数に分解する:

169 = 1 3^{2}

= 1 3^{2}

累乗根の規則を適用する:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 21}

解を分離する

u_{1} = \frac{- 1 + 13}{2 \cdot 21}, u_{2} = \frac{- 1 - 13}{2 \cdot 21}

u = \frac{- 1 + 13}{2 \cdot 21} : \frac{2}{7}

\frac{- 1 + 13}{2 \cdot 21}

数を足す/引く:

- 1 + 13 = 12

= \frac{12}{2 \cdot 21}

数を乗じる:

2 \cdot 21 = 42

= \frac{12}{42}

共通因数を約分する:

6

= \frac{2}{7}

u = \frac{- 1 - 13}{2 \cdot 21} : - \frac{1}{3}

\frac{- 1 - 13}{2 \cdot 21}

数を引く:

- 1 - 13 = - 14

= \frac{- 14}{2 \cdot 21}

数を乗じる:

2 \cdot 21 = 42

= \frac{- 14}{42}

分数の規則を適用する:

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

= - \frac{14}{42}

共通因数を約分する:

14

= - \frac{1}{3}

二次equationの解:

u = \frac{2}{7}, u = - \frac{1}{3}

代用を戻す

u = cos (x)

cos (x) = \frac{2}{7}, cos (x) = - \frac{1}{3}

cos (x) = \frac{2}{7}, cos (x) = - \frac{1}{3}

cos (x) = \frac{2}{7} : x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

cos (x) = \frac{2}{7}

三角関数の逆数プロパティを適用する

cos (x) = \frac{2}{7}

以下の一般解

cos (x) = \frac{2}{7}

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

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

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

すべての解を組み合わせる

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn, x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

10進法形式で解を証明する

x = 1.28104 \dots + 2 πn, x = 2 π - 1.28104 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

グラフ

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