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人気のある三角関数 >

2cos^2(x)(1+2cos^2(x))=2

解

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

解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

2 u^{2} (1 + 2 u^{2}) = 2

2 u^{2} (1 + 2 u^{2}) = 2 : u = \frac{1}{2}, u = - \frac{1}{2}, u = i, u = - i

2 u^{2} (1 + 2 u^{2}) = 2

拡張

2 u^{2} (1 + 2 u^{2}) : 2 u^{2} + 4 u^{4}

2 u^{2} (1 + 2 u^{2})

分配法則を適用する:

a (b + c) = ab + a c

a = 2 u^{2}, b = 1, c = 2 u^{2}

= 2 u^{2} \cdot 1 + 2 u^{2} \cdot 2 u^{2}

= 2 \cdot 1 \cdot u^{2} + 2 \cdot 2 u^{2} u^{2}

簡素化

2 \cdot 1 \cdot u^{2} + 2 \cdot 2 u^{2} u^{2} : 2 u^{2} + 4 u^{4}

2 \cdot 1 \cdot u^{2} + 2 \cdot 2 u^{2} u^{2}

2 \cdot 1 \cdot u^{2} = 2 u^{2}

2 \cdot 1 \cdot u^{2}

数を乗じる:

2 \cdot 1 = 2

= 2 u^{2}

2 \cdot 2 u^{2} u^{2} = 4 u^{4}

2 \cdot 2 u^{2} u^{2}

数を乗じる:

2 \cdot 2 = 4

= 4 u^{2} u^{2}

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= 4 u^{2 + 2}

数を足す:

2 + 2 = 4

= 4 u^{4}

= 2 u^{2} + 4 u^{4}

= 2 u^{2} + 4 u^{4}

2 u^{2} + 4 u^{4} = 2

2

を左側に移動します

2 u^{2} + 4 u^{4} = 2

両辺から

2

を引く

2 u^{2} + 4 u^{4} - 2 = 2 - 2

簡素化

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

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

標準的な形式で書く

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

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

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

4 v^{2} + 2 v - 2 = 0

解く

4 v^{2} + 2 v - 2 = 0 : v = \frac{1}{2}, v = - 1

4 v^{2} + 2 v - 2 = 0

解くとthe二次式

4 v^{2} + 2 v - 2 = 0

二次Equationの公式:

次の場合:

a = 4, b = 2, c = - 2

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

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

2^{2} - 4 \cdot 4 (- 2) = 6

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

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 2

数を乗じる:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

v_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

解を分離する

v_{1} = \frac{- 2 + 6}{2 \cdot 4}, v_{2} = \frac{- 2 - 6}{2 \cdot 4}

v = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

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

数を足す/引く:

- 2 + 6 = 4

= \frac{4}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{4}{8}

共通因数を約分する:

4

= \frac{1}{2}

v = \frac{- 2 - 6}{2 \cdot 4} : - 1

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

数を引く:

- 2 - 6 = - 8

= \frac{- 8}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 8}{8}

分数の規則を適用する:

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

= - \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

v = \frac{1}{2}, v = - 1

v = \frac{1}{2}, v = - 1

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

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

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

x^{2} = f (a)

の場合, 解は

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

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

解く

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

x^{2} = f (a)

の場合, 解は

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

u = - 1, u = - - 1

簡素化

- 1 : i

- 1

虚数の規則を適用する:

- 1 = i

= i

簡素化

- - 1 : - i

- - 1

虚数の規則を適用する:

- 1 = i

= - i

u = i, u = - i

解答は

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

代用を戻す

u = cos (x)

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

cos (x) = i :

解なし

cos (x) = i

解なし

cos (x) = - i :

解なし

cos (x) = - i

解なし

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

グラフ

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