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人気のある三角関数 >

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

解

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

解

x = 0.46364 \dots + 2 πn, x = 2 π - 0.46364 \dots + 2 πn, x = 2.67794 \dots + 2 πn, x = - 2.67794 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度

x = 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 333.43494 \dots^{\circ} + 36 0^{\circ} n, x = 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = - 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

三角関数の公式を使用して書き換える

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

2倍角の公式を使用:

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

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

簡素化

- 5 + 4 cos^{2} (x) + 5 (2 cos^{2} (x) - 1)^{2} : 20 cos^{4} (x) - 16 cos^{2} (x)

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

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

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

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

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

簡素化

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

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

(cos^{2} (x))^{2} : cos^{4} (x)

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

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

数を乗じる:

2 \cdot 2 = 4

= cos^{4} (x)

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

2^{2} = 4

= 4 cos^{4} (x)

2 \cdot 2 cos^{2} (x) \cdot 1 = 4 cos^{2} (x)

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

数を乗じる:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (x)

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

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

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

拡張

5 (4 cos^{4} (x) - 4 cos^{2} (x) + 1) : 20 cos^{4} (x) - 20 cos^{2} (x) + 5

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

括弧を分配する

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

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

5 \cdot 4 cos^{4} (x) - 5 \cdot 4 cos^{2} (x) + 5 \cdot 1 : 20 cos^{4} (x) - 20 cos^{2} (x) + 5

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

数を乗じる:

5 \cdot 4 = 20

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5 \cdot 1

数を乗じる:

5 \cdot 1 = 5

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5

= 20 cos^{4} (x) - 20 cos^{2} (x) + 5

= - 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5

簡素化

- 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5 : 20 cos^{4} (x) - 16 cos^{2} (x)

- 5 + 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) + 5

条件のようなグループ

= 4 cos^{2} (x) + 20 cos^{4} (x) - 20 cos^{2} (x) - 5 + 5

類似した元を足す:

4 cos^{2} (x) - 20 cos^{2} (x) = - 16 cos^{2} (x)

= - 16 cos^{2} (x) + 20 cos^{4} (x) - 5 + 5

- 5 + 5 = 0

= 20 cos^{4} (x) - 16 cos^{2} (x)

= 20 cos^{4} (x) - 16 cos^{2} (x)

= 20 cos^{4} (x) - 16 cos^{2} (x)

- 16 cos^{2} (x) + 20 cos^{4} (x) = 0

置換で解く

- 16 cos^{2} (x) + 20 cos^{4} (x) = 0

仮定:

cos (x) = u

- 16 u^{2} + 20 u^{4} = 0

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

- 16 u^{2} + 20 u^{4} = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

20 u^{4} - 16 u^{2} = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

20 v^{2} - 16 v = 0

解く

20 v^{2} - 16 v = 0 : v = \frac{4}{5}, v = 0

20 v^{2} - 16 v = 0

解くとthe二次式

20 v^{2} - 16 v = 0

二次Equationの公式:

次の場合:

a = 20, b = - 16, c = 0

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 20 \cdot 0}{2 \cdot 20}

v_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 20 \cdot 0}{2 \cdot 20}

(- 16)^{2} - 4 \cdot 20 \cdot 0 = 16

(- 16)^{2} - 4 \cdot 20 \cdot 0

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 20 \cdot 0

規則を適用

0 \cdot a = 0

= 1 6^{2} - 0

1 6^{2} - 0 = 1 6^{2}

= 1 6^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 16

v_{1, 2} = \frac{- ( - 16 ) \pm 16}{2 \cdot 20}

解を分離する

v_{1} = \frac{- ( - 16 ) + 16}{2 \cdot 20}, v_{2} = \frac{- ( - 16 ) - 16}{2 \cdot 20}

v = \frac{- ( - 16 ) + 16}{2 \cdot 20} : \frac{4}{5}

\frac{- ( - 16 ) + 16}{2 \cdot 20}

規則を適用

- (- a) = a

= \frac{16 + 16}{2 \cdot 20}

数を足す:

16 + 16 = 32

= \frac{32}{2 \cdot 20}

数を乗じる:

2 \cdot 20 = 40

= \frac{32}{40}

共通因数を約分する:

8

= \frac{4}{5}

v = \frac{- ( - 16 ) - 16}{2 \cdot 20} : 0

\frac{- ( - 16 ) - 16}{2 \cdot 20}

規則を適用

- (- a) = a

= \frac{16 - 16}{2 \cdot 20}

数を引く:

16 - 16 = 0

= \frac{0}{2 \cdot 20}

数を乗じる:

2 \cdot 20 = 40

= \frac{0}{40}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

二次equationの解:

v = \frac{4}{5}, v = 0

v = \frac{4}{5}, v = 0

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{4}{5} : u = \frac{2 5}{5}, u = - \frac{2 5}{5}

u^{2} = \frac{4}{5}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

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

\frac{4}{5} = \frac{2 5}{5}

\frac{4}{5}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{4}{5}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{5}

有理化する

\frac{2}{5} : \frac{2 5}{5}

\frac{2}{5}

共役で乗じる

\frac{5}{5}

= \frac{2 5}{5 5}

55 = 5

55

累乗根の規則を適用する:

a a = a

55 = 5

= 5

= \frac{2 5}{5}

= \frac{2 5}{5}

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

- \frac{4}{5}

簡素化

\frac{4}{5} : \frac{2}{5}

\frac{4}{5}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{4}{5}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{5}

= - \frac{2}{5}

有理化する

- \frac{2}{5} : - \frac{2 5}{5}

- \frac{2}{5}

共役で乗じる

\frac{5}{5}

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

55 = 5

55

累乗根の規則を適用する:

a a = a

55 = 5

= 5

= - \frac{2 5}{5}

= - \frac{2 5}{5}

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

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

解答は

u = \frac{2 5}{5}, u = - \frac{2 5}{5}, u = 0

代用を戻す

u = cos (x)

cos (x) = \frac{2 5}{5}, cos (x) = - \frac{2 5}{5}, cos (x) = 0

cos (x) = \frac{2 5}{5}, cos (x) = - \frac{2 5}{5}, cos (x) = 0

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

cos (x) = \frac{2 5}{5}

三角関数の逆数プロパティを適用する

cos (x) = \frac{2 5}{5}

以下の一般解

cos (x) = \frac{2 5}{5}

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

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

以下の一般解

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.46364 \dots + 2 πn, x = 2 π - 0.46364 \dots + 2 πn, x = 2.67794 \dots + 2 πn, x = - 2.67794 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

グラフ

人気の例

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sin (3 x + 10) = cos (x + 2 4^{\circ})

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\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}

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d^{2} + 13 d + 36 = \frac{sin ^{2} ( x )}{2}

(tan^2(x)-4)/(cos(x)+5)=0

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