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人気のある三角関数 >

5sin^2(x)+6cos(x)-6=0

解

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

解

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

+1

度

x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

三角関数の公式を使用して書き換える

- 6 + 5 sin^{2} (x) + 6 cos (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)

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

簡素化

- 6 + 5 (1 - cos^{2} (x)) + 6 cos (x) : 6 cos (x) - 5 cos^{2} (x) - 1

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

拡張

5 (1 - cos^{2} (x)) : 5 - 5 cos^{2} (x)

5 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 5, b = 1, c = cos^{2} (x)

= 5 \cdot 1 - 5 cos^{2} (x)

数を乗じる:

5 \cdot 1 = 5

= 5 - 5 cos^{2} (x)

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

数を足す/引く:

- 6 + 5 = - 1

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 5, b = 6, c = - 1

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

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

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

6^{2} - 4 (- 5) (- 1)

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 5 \cdot 1 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

数を引く:

36 - 20 = 16

= 16

数を因数に分解する:

16 = 4^{2}

= 4^{2}

累乗根の規則を適用する:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- 6 \pm 4}{2 ( - 5 )}

解を分離する

u_{1} = \frac{- 6 + 4}{2 ( - 5 )}, u_{2} = \frac{- 6 - 4}{2 ( - 5 )}

u = \frac{- 6 + 4}{2 ( - 5 )} : \frac{1}{5}

\frac{- 6 + 4}{2 ( - 5 )}

括弧を削除する:

(- a) = - a

= \frac{- 6 + 4}{- 2 \cdot 5}

数を足す/引く:

- 6 + 4 = - 2

= \frac{- 2}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{- 2}{- 10}

分数の規則を適用する:

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

= \frac{2}{10}

共通因数を約分する:

2

= \frac{1}{5}

u = \frac{- 6 - 4}{2 ( - 5 )} : 1

\frac{- 6 - 4}{2 ( - 5 )}

括弧を削除する:

(- a) = - a

= \frac{- 6 - 4}{- 2 \cdot 5}

数を引く:

- 6 - 4 = - 10

= \frac{- 10}{- 2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{- 10}{- 10}

分数の規則を適用する:

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

= \frac{10}{10}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

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

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

以下の一般解

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

グラフ

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