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人気のある三角関数 >

20sin^2(x)+cos(x)-19=0

解

20 sin^{2} (x) + cos (x) - 19 = 0

解

x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

+1

度

x = 101.53695 \dots^{\circ} + 36 0^{\circ} n, x = - 101.53695 \dots^{\circ} + 36 0^{\circ} n, x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

20 sin^{2} (x) + cos (x) - 19 = 0

三角関数の公式を使用して書き換える

- 19 + cos (x) + 20 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)

= - 19 + cos (x) + 20 (1 - cos^{2} (x))

簡素化

- 19 + cos (x) + 20 (1 - cos^{2} (x)) : cos (x) - 20 cos^{2} (x) + 1

- 19 + cos (x) + 20 (1 - cos^{2} (x))

拡張

20 (1 - cos^{2} (x)) : 20 - 20 cos^{2} (x)

20 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 20, b = 1, c = cos^{2} (x)

= 20 \cdot 1 - 20 cos^{2} (x)

数を乗じる:

20 \cdot 1 = 20

= 20 - 20 cos^{2} (x)

= - 19 + cos (x) + 20 - 20 cos^{2} (x)

簡素化

- 19 + cos (x) + 20 - 20 cos^{2} (x) : cos (x) - 20 cos^{2} (x) + 1

- 19 + cos (x) + 20 - 20 cos^{2} (x)

条件のようなグループ

= cos (x) - 20 cos^{2} (x) - 19 + 20

数を足す/引く:

- 19 + 20 = 1

= cos (x) - 20 cos^{2} (x) + 1

= cos (x) - 20 cos^{2} (x) + 1

= cos (x) - 20 cos^{2} (x) + 1

1 + cos (x) - 20 cos^{2} (x) = 0

置換で解く

1 + cos (x) - 20 cos^{2} (x) = 0

仮定:

cos (x) = u

1 + u - 20 u^{2} = 0

1 + u - 20 u^{2} = 0 : u = - \frac{1}{5}, u = \frac{1}{4}

1 + u - 20 u^{2} = 0

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 20, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 20 ) \cdot 1}{2 ( - 20 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 20 ) \cdot 1}{2 ( - 20 )}

1^{2} - 4 (- 20) \cdot 1 = 9

1^{2} - 4 (- 20) \cdot 1

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 20) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 20 \cdot 1

数を乗じる:

4 \cdot 20 \cdot 1 = 80

= 1 + 80

数を足す:

1 + 80 = 81

= 81

数を因数に分解する:

81 = 9^{2}

= 9^{2}

累乗根の規則を適用する:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- 1 \pm 9}{2 ( - 20 )}

解を分離する

u_{1} = \frac{- 1 + 9}{2 ( - 20 )}, u_{2} = \frac{- 1 - 9}{2 ( - 20 )}

u = \frac{- 1 + 9}{2 ( - 20 )} : - \frac{1}{5}

\frac{- 1 + 9}{2 ( - 20 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 9}{- 2 \cdot 20}

数を足す/引く:

- 1 + 9 = 8

= \frac{8}{- 2 \cdot 20}

数を乗じる:

2 \cdot 20 = 40

= \frac{8}{- 40}

分数の規則を適用する:

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

= - \frac{8}{40}

共通因数を約分する:

8

= - \frac{1}{5}

u = \frac{- 1 - 9}{2 ( - 20 )} : \frac{1}{4}

\frac{- 1 - 9}{2 ( - 20 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 9}{- 2 \cdot 20}

数を引く:

- 1 - 9 = - 10

= \frac{- 10}{- 2 \cdot 20}

数を乗じる:

2 \cdot 20 = 40

= \frac{- 10}{- 40}

分数の規則を適用する:

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

= \frac{10}{40}

共通因数を約分する:

10

= \frac{1}{4}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

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

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

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

すべての解を組み合わせる

x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn, x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

10進法形式で解を証明する

x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

グラフ

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