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人気のある三角関数 >

5cos^2(x)= 1/(2(1+cos^2(x)))

解

5 cos^{2} (x) = \frac{1}{2 ( 1 + cos ^{2} ( x ) )}

解

x = 1.26330 \dots + 2 πn, x = 2 π - 1.26330 \dots + 2 πn, x = 1.87828 \dots + 2 πn, x = - 1.87828 \dots + 2 πn

+1

度

x = 72.38207 \dots^{\circ} + 36 0^{\circ} n, x = 287.61792 \dots^{\circ} + 36 0^{\circ} n, x = 107.61792 \dots^{\circ} + 36 0^{\circ} n, x = - 107.61792 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

5 cos^{2} (x) = \frac{1}{2 ( 1 + cos ^{2} ( x ) )}

置換で解く

5 cos^{2} (x) = \frac{1}{2 ( 1 + cos ^{2} ( x ) )}

仮定:

cos (x) = u

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )} : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

以下で両辺を乗じる:

2 (1 + u^{2})

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

以下で両辺を乗じる:

2 (1 + u^{2})

5 u^{2} \cdot 2 (1 + u^{2}) = \frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2})

簡素化

5 u^{2} \cdot 2 (1 + u^{2}) = \frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2})

簡素化

5 u^{2} \cdot 2 (1 + u^{2}) : 10 u^{2} (1 + u^{2})

5 u^{2} \cdot 2 (1 + u^{2})

数を乗じる:

5 \cdot 2 = 10

= 10 u^{2} (u^{2} + 1)

簡素化

\frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2}) : 1

\frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2})

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 ( 1 + u ^{2} )}{2 ( 1 + u ^{2} )}

共通因数を約分する:

2

= \frac{1 \cdot ( 1 + u ^{2} )}{1 + u ^{2}}

共通因数を約分する:

1 + u^{2}

= 1

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

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

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

解く

10 u^{2} (1 + u^{2}) = 1 : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

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

拡張

10 u^{2} (1 + u^{2}) : 10 u^{2} + 10 u^{4}

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

分配法則を適用する:

a (b + c) = ab + a c

a = 10 u^{2}, b = 1, c = u^{2}

= 10 u^{2} \cdot 1 + 10 u^{2} u^{2}

= 10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2}

簡素化

10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2} : 10 u^{2} + 10 u^{4}

10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2}

10 \cdot 1 \cdot u^{2} = 10 u^{2}

10 \cdot 1 \cdot u^{2}

数を乗じる:

10 \cdot 1 = 10

= 10 u^{2}

10 u^{2} u^{2} = 10 u^{4}

10 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}

= 10 u^{2 + 2}

数を足す:

2 + 2 = 4

= 10 u^{4}

= 10 u^{2} + 10 u^{4}

= 10 u^{2} + 10 u^{4}

10 u^{2} + 10 u^{4} = 1

1

を左側に移動します

10 u^{2} + 10 u^{4} = 1

両辺から

1

を引く

10 u^{2} + 10 u^{4} - 1 = 1 - 1

簡素化

10 u^{2} + 10 u^{4} - 1 = 0

10 u^{2} + 10 u^{4} - 1 = 0

標準的な形式で書く

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

10 u^{4} + 10 u^{2} - 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

10 v^{2} + 10 v - 1 = 0

解く

10 v^{2} + 10 v - 1 = 0 : v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

10 v^{2} + 10 v - 1 = 0

解くとthe二次式

10 v^{2} + 10 v - 1 = 0

二次Equationの公式:

次の場合:

a = 10, b = 10, c = - 1

v_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

v_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

1 0^{2} - 4 \cdot 10 (- 1) = 235

1 0^{2} - 4 \cdot 10 (- 1)

規則を適用

- (- a) = a

= 1 0^{2} + 4 \cdot 10 \cdot 1

数を乗じる:

4 \cdot 10 \cdot 1 = 40

= 1 0^{2} + 40

1 0^{2} = 100

= 100 + 40

数を足す:

100 + 40 = 140

= 140

以下の素因数分解:

140 : 2^{2} \cdot 5 \cdot 7

140

1402140 = 70 \cdot 2

で割る

= 2 \cdot 70

70270 = 35 \cdot 2

で割る

= 2 \cdot 2 \cdot 35

35535 = 7 \cdot 5

で割る

= 2 \cdot 2 \cdot 5 \cdot 7

2, 5, 7

はすべて素数である。ゆえにさらに因数分解することはできない

= 2 \cdot 2 \cdot 5 \cdot 7

= 2^{2} \cdot 5 \cdot 7

= 2^{2} \cdot 5 \cdot 7

累乗根の規則を適用する:

n ab = n a n b

= 2^{2} 5 \cdot 7

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2 5 \cdot 7

改良

= 235

v_{1, 2} = \frac{- 10 \pm 2 35}{2 \cdot 10}

解を分離する

v_{1} = \frac{- 10 + 2 35}{2 \cdot 10}, v_{2} = \frac{- 10 - 2 35}{2 \cdot 10}

v = \frac{- 10 + 2 35}{2 \cdot 10} : \frac{- 5 + 35}{10}

\frac{- 10 + 2 35}{2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{- 10 + 2 35}{20}

因数

- 10 + 235 : 2 (- 5 + 35)

- 10 + 235

書き換え

= - 2 \cdot 5 + 235

共通項をくくり出す

2

= 2 (- 5 + 35)

= \frac{2 ( - 5 + 35 )}{20}

共通因数を約分する:

2

= \frac{- 5 + 35}{10}

v = \frac{- 10 - 2 35}{2 \cdot 10} : - \frac{5 + 35}{10}

\frac{- 10 - 2 35}{2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{- 10 - 2 35}{20}

因数

- 10 - 235 : - 2 (5 + 35)

- 10 - 235

書き換え

= - 2 \cdot 5 - 235

共通項をくくり出す

2

= - 2 (5 + 35)

= - \frac{2 ( 5 + 35 )}{20}

共通因数を約分する:

2

= - \frac{5 + 35}{10}

二次equationの解:

v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{- 5 + 35}{10} : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}

u^{2} = \frac{- 5 + 35}{10}

x^{2} = f (a)

の場合, 解は

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

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}

解く

u^{2} = - \frac{5 + 35}{10} : u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

u^{2} = - \frac{5 + 35}{10}

x^{2} = f (a)

の場合, 解は

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

u = - \frac{5 + 35}{10}, u = - - \frac{5 + 35}{10}

簡素化

- \frac{5 + 35}{10} : i \frac{5 + 35}{10}

- \frac{5 + 35}{10}

累乗根の規則を適用する:

- a = - 1 a

- \frac{5 + 35}{10} = - 1 \frac{5 + 35}{10}

= - 1 \frac{5 + 35}{10}

虚数の規則を適用する:

- 1 = i

= i \frac{5 + 35}{10}

簡素化

- - \frac{5 + 35}{10} : - i \frac{5 + 35}{10}

- - \frac{5 + 35}{10}

簡素化

- \frac{5 + 35}{10} : i \frac{5 + 35}{10}

- \frac{5 + 35}{10}

累乗根の規則を適用する:

- a = - 1 a

- \frac{5 + 35}{10} = - 1 \frac{5 + 35}{10}

= - 1 \frac{5 + 35}{10}

虚数の規則を適用する:

- 1 = i

= i \frac{5 + 35}{10}

= - i \frac{5 + 35}{10}

u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

解答は

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

代用を戻す

u = cos (x)

cos (x) = \frac{- 5 + 35}{10}, cos (x) = - \frac{- 5 + 35}{10}, cos (x) = i \frac{5 + 35}{10}, cos (x) = - i \frac{5 + 35}{10}

cos (x) = \frac{- 5 + 35}{10}, cos (x) = - \frac{- 5 + 35}{10}, cos (x) = i \frac{5 + 35}{10}, cos (x) = - i \frac{5 + 35}{10}

cos (x) = \frac{- 5 + 35}{10} : x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

cos (x) = \frac{- 5 + 35}{10}

三角関数の逆数プロパティを適用する

cos (x) = \frac{- 5 + 35}{10}

以下の一般解

cos (x) = \frac{- 5 + 35}{10}

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

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

cos (x) = - \frac{- 5 + 35}{10} : x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

cos (x) = - \frac{- 5 + 35}{10}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{- 5 + 35}{10}

以下の一般解

cos (x) = - \frac{- 5 + 35}{10}

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

x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

cos (x) = i \frac{5 + 35}{10} :

解なし

cos (x) = i \frac{5 + 35}{10}

解なし

cos (x) = - i \frac{5 + 35}{10} :

解なし

cos (x) = - i \frac{5 + 35}{10}

解なし

すべての解を組み合わせる

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn, x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

10進法形式で解を証明する

x = 1.26330 \dots + 2 πn, x = 2 π - 1.26330 \dots + 2 πn, x = 1.87828 \dots + 2 πn, x = - 1.87828 \dots + 2 πn

グラフ

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