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人気のある三角関数 >

solvefor x,f[g]=cos(1/(x^2))

解

解く

x, f [g] = cos (\frac{1}{x ^{2}})

解

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

解答ステップ

f [g] = cos (\frac{1}{x ^{2}})

辺を交換する

cos (\frac{1}{x ^{2}}) = f [g]

括弧を削除する:

(a) = a

cos (\frac{1}{x ^{2}}) = f g

三角関数の逆数プロパティを適用する

cos (\frac{1}{x ^{2}}) = f g

以下の一般解

cos (\frac{1}{x ^{2}}) = f g

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

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

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

解く

\frac{1}{x ^{2}} = arccos (f g) + 2 πn : x = \frac{1}{arccos ( f g ) + 2 πn}, x = - \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

\frac{1}{x ^{2}} = arccos (f g) + 2 πn

以下で両辺を乗じる:

x^{2}

\frac{1}{x ^{2}} = arccos (f g) + 2 πn

以下で両辺を乗じる:

x^{2}

\frac{1}{x ^{2}} x^{2} = arccos (f g) x^{2} + 2 πn x^{2}

簡素化

1 = arccos (f g) x^{2} + 2 πn x^{2}

1 = arccos (f g) x^{2} + 2 πn x^{2}

解く

1 = arccos (f g) x^{2} + 2 πn x^{2} : x = \frac{1}{arccos ( f g ) + 2 πn}, x = - \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

1 = arccos (f g) x^{2} + 2 πn x^{2}

辺を交換する

arccos (f g) x^{2} + 2 πn x^{2} = 1

因数

arccos (f g) x^{2} + 2 πn x^{2} : x^{2} (arccos (f g) + 2 πn)

arccos (f g) x^{2} + 2 πn x^{2}

共通項をくくり出す

x^{2}

= x^{2} (arccos (f g) + 2 πn)

x^{2} (arccos (f g) + 2 πn) = 1

以下で両辺を割る

arccos (f g) + 2 πn; f \neq = \frac{cos ( - 2 πn )}{g}

x^{2} (arccos (f g) + 2 πn) = 1

以下で両辺を割る

arccos (f g) + 2 πn; f \neq = \frac{cos ( - 2 πn )}{g}

\frac{x ^{2} ( arccos ( f g ) + 2 πn )}{arccos ( f g ) + 2 πn} = \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

簡素化

x^{2} = \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

x^{2} = \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

x^{2} = f (a)

の場合, 解は

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

x = \frac{1}{arccos ( f g ) + 2 πn}, x = - \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

x = \frac{1}{arccos ( f g ) + 2 πn}, x = - \frac{1}{arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( - 2 πn )}{g}

解く

\frac{1}{x ^{2}} = - arccos (f g) + 2 πn : x = \frac{1}{- arccos ( f g ) + 2 πn}, x = - \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

\frac{1}{x ^{2}} = - arccos (f g) + 2 πn

以下で両辺を乗じる:

x^{2}

\frac{1}{x ^{2}} = - arccos (f g) + 2 πn

以下で両辺を乗じる:

x^{2}

\frac{1}{x ^{2}} x^{2} = - arccos (f g) x^{2} + 2 πn x^{2}

簡素化

1 = - arccos (f g) x^{2} + 2 πn x^{2}

1 = - arccos (f g) x^{2} + 2 πn x^{2}

解く

1 = - arccos (f g) x^{2} + 2 πn x^{2} : x = \frac{1}{- arccos ( f g ) + 2 πn}, x = - \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

1 = - arccos (f g) x^{2} + 2 πn x^{2}

辺を交換する

- arccos (f g) x^{2} + 2 πn x^{2} = 1

因数

- arccos (f g) x^{2} + 2 πn x^{2} : x^{2} (- arccos (f g) + 2 πn)

- arccos (f g) x^{2} + 2 πn x^{2}

共通項をくくり出す

x^{2}

= x^{2} (- arccos (f g) + 2 πn)

x^{2} (- arccos (f g) + 2 πn) = 1

以下で両辺を割る

- arccos (f g) + 2 πn; f \neq = \frac{cos ( 2 πn )}{g}

x^{2} (- arccos (f g) + 2 πn) = 1

以下で両辺を割る

- arccos (f g) + 2 πn; f \neq = \frac{cos ( 2 πn )}{g}

\frac{x ^{2} ( - arccos ( f g ) + 2 πn )}{- arccos ( f g ) + 2 πn} = \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

簡素化

x^{2} = \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

x^{2} = \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

x^{2} = f (a)

の場合, 解は

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

x = \frac{1}{- arccos ( f g ) + 2 πn}, x = - \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

x = \frac{1}{- arccos ( f g ) + 2 πn}, x = - \frac{1}{- arccos ( f g ) + 2 πn}; f \neq = \frac{cos ( 2 πn )}{g}

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

グラフ

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