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人気のある三角関数 >

5cos(x)=-2sin^2(x)+4,0<= x<= 2pi

解

5 cos (x) = - 2 sin^{2} (x) + 4, 0 \leq x \leq 2 π

解

x = \frac{π}{3}, x = \frac{5 π}{3}

+1

度

x = 6 0^{\circ}, x = 30 0^{\circ}

解答ステップ

5 cos (x) = - 2 sin^{2} (x) + 4, 0 \leq x \leq 2 π

両辺から

- 2 sin^{2} (x) + 4

を引く

5 cos (x) + 2 sin^{2} (x) - 4 = 0

三角関数の公式を使用して書き換える

- 4 + 2 sin^{2} (x) + 5 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)

= - 4 + 2 (1 - cos^{2} (x)) + 5 cos (x)

簡素化

- 4 + 2 (1 - cos^{2} (x)) + 5 cos (x) : 5 cos (x) - 2 cos^{2} (x) - 2

- 4 + 2 (1 - cos^{2} (x)) + 5 cos (x)

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

= - 4 + 2 - 2 cos^{2} (x) + 5 cos (x)

数を足す/引く:

- 4 + 2 = - 2

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = 5, c = - 2

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

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

5^{2} - 4 (- 2) (- 2) = 3

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

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数を引く:

25 - 16 = 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{- 5 \pm 3}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 5 + 3}{2 ( - 2 )}, u_{2} = \frac{- 5 - 3}{2 ( - 2 )}

u = \frac{- 5 + 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- 5 + 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 + 3}{- 2 \cdot 2}

数を足す/引く:

- 5 + 3 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 5 - 3}{2 ( - 2 )} : 2

\frac{- 5 - 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 - 3}{- 2 \cdot 2}

数を引く:

- 5 - 3 = - 8

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 8}{- 4}

分数の規則を適用する:

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

= \frac{8}{4}

数を割る:

\frac{8}{4} = 2

= 2

二次equationの解:

u = \frac{1}{2}, u = 2

代用を戻す

u = cos (x)

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

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

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = 2, 0 \leq x \leq 2 π :

解なし

cos (x) = 2, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

x = \frac{π}{3}, x = \frac{5 π}{3}

グラフ

人気の例

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\frac{1 + tan ^{2} ( x )}{1 + tan ^{2} ( x )} = 1 - sin^{2} (x)

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