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人気のある三角関数 >

cos(pi/x)=(sqrt(3))/3

解

cos (\frac{π}{x}) = \frac{3}{3}

解

x = \frac{π}{0.95531 \dots + 2 πn}, x = \frac{π}{2 π - 0.95531 \dots + 2 πn}

+1

度

x = 0^{\circ} + 24.86702 \dots^{\circ} n, x = 0^{\circ} + 15.50246 \dots^{\circ} n

解答ステップ

cos (\frac{π}{x}) = \frac{3}{3}

三角関数の逆数プロパティを適用する

cos (\frac{π}{x}) = \frac{3}{3}

以下の一般解

cos (\frac{π}{x}) = \frac{3}{3}

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

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

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

解く

\frac{π}{x} = arccos (\frac{3}{3}) + 2 πn : x = \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

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

以下で両辺を乗じる:

x

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

以下で両辺を乗じる:

x

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

簡素化

π = arccos (\frac{3}{3}) x + 2 πn x

π = arccos (\frac{3}{3}) x + 2 πn x

辺を交換する

arccos (\frac{3}{3}) x + 2 πn x = π

簡素化

arccos (\frac{3}{3}) x + 2 πn x : arccos (\frac{1}{3}) x + 2 πn x

arccos (\frac{3}{3}) x + 2 πn x

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= arccos (\frac{1}{3}) x + 2 πn x

arccos (\frac{1}{3}) x + 2 πn x = π

arccos (\frac{3}{3}) x + 2 πn x = π

因数

arccos (\frac{3}{3}) x + 2 πn x : x (arccos (\frac{1}{3}) + 2 πn)

arccos (\frac{3}{3}) x + 2 πn x

共通項をくくり出す

x

= x (arccos (\frac{3}{3}) + 2 πn)

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= x (2 πn + arccos (\frac{1}{3}))

x (arccos (\frac{1}{3}) + 2 πn) = π

以下で両辺を割る

arccos (\frac{1}{3}) + 2 πn; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

x (arccos (\frac{1}{3}) + 2 πn) = π

以下で両辺を割る

arccos (\frac{1}{3}) + 2 πn; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

\frac{x ( arccos ( \frac{1}{3} ) + 2 πn )}{arccos ( \frac{1}{3} ) + 2 πn} = \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

簡素化

\frac{x ( arccos ( \frac{1}{3} ) + 2 πn )}{arccos ( \frac{1}{3} ) + 2 πn} = \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}

簡素化

\frac{x ( arccos ( \frac{1}{3} ) + 2 πn )}{arccos ( \frac{1}{3} ) + 2 πn} : x

\frac{x ( arccos ( \frac{1}{3} ) + 2 πn )}{arccos ( \frac{1}{3} ) + 2 πn}

x (arccos (\frac{1}{3}) + 2 πn) = x (arccos (\frac{3}{3}) + 2 πn)

x (arccos (\frac{1}{3}) + 2 πn)

= x (2 πn + arccos (\frac{3}{3}))

= \frac{x ( 2 πn + arccos ( \frac{3}{3} ) )}{arccos ( \frac{1}{3} ) + 2 πn}

arccos (\frac{1}{3}) + 2 πn = arccos (\frac{3}{3}) + 2 πn

arccos (\frac{1}{3}) + 2 πn

= arccos (\frac{3}{3}) + 2 πn

= \frac{x ( 2 πn + arccos ( \frac{3}{3} ) )}{arccos ( \frac{3}{3} ) + 2 πn}

共通因数を約分する:

arccos (\frac{3}{3}) + 2 πn

= x

簡素化

\frac{π}{arccos ( \frac{1}{3} ) + 2 πn} : \frac{π}{arccos ( \frac{3}{3} ) + 2 πn}

\frac{π}{arccos ( \frac{1}{3} ) + 2 πn}

arccos (\frac{1}{3}) + 2 πn = arccos (\frac{3}{3}) + 2 πn

arccos (\frac{1}{3}) + 2 πn

= arccos (\frac{3}{3}) + 2 πn

= \frac{π}{arccos ( \frac{3}{3} ) + 2 πn}

x = \frac{π}{arccos ( \frac{3}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

x = \frac{π}{arccos ( \frac{3}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

x = \frac{π}{arccos ( \frac{3}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

簡素化

\frac{π}{arccos ( \frac{3}{3} ) + 2 πn} : \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}

\frac{π}{arccos ( \frac{3}{3} ) + 2 πn}

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}

x = \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}; n \neq = - \frac{arccos ( \frac{1}{3} )}{2 π}

解く

\frac{π}{x} = 2 π - arccos (\frac{3}{3}) + 2 πn : x = \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

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

以下で両辺を乗じる:

x

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

以下で両辺を乗じる:

x

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

簡素化

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

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

辺を交換する

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

簡素化

2 π x - arccos (\frac{3}{3}) x + 2 πn x : 2 π x - arccos (\frac{1}{3}) x + 2 πn x

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

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= 2 π x - arccos (\frac{1}{3}) x + 2 πn x

2 π x - arccos (\frac{1}{3}) x + 2 πn x = π

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

因数

2 π x - arccos (\frac{3}{3}) x + 2 πn x : x (2 π - arccos (\frac{1}{3}) + 2 πn)

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

共通項をくくり出す

x

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

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= x (2 π + 2 πn - arccos (\frac{1}{3}))

x (2 π - arccos (\frac{1}{3}) + 2 πn) = π

以下で両辺を割る

2 π - arccos (\frac{1}{3}) + 2 πn; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

x (2 π - arccos (\frac{1}{3}) + 2 πn) = π

以下で両辺を割る

2 π - arccos (\frac{1}{3}) + 2 πn; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

\frac{x ( 2 π - arccos ( \frac{1}{3} ) + 2 πn )}{2 π - arccos ( \frac{1}{3} ) + 2 πn} = \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

簡素化

\frac{x ( 2 π - arccos ( \frac{1}{3} ) + 2 πn )}{2 π - arccos ( \frac{1}{3} ) + 2 πn} = \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

簡素化

\frac{x ( 2 π - arccos ( \frac{1}{3} ) + 2 πn )}{2 π - arccos ( \frac{1}{3} ) + 2 πn} : x

\frac{x ( 2 π - arccos ( \frac{1}{3} ) + 2 πn )}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

x (2 π - arccos (\frac{1}{3}) + 2 πn) = x (2 π - arccos (\frac{3}{3}) + 2 πn)

x (2 π - arccos (\frac{1}{3}) + 2 πn)

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

= \frac{x ( 2 π + 2 πn - arccos ( \frac{3}{3} ) )}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

2 π - arccos (\frac{1}{3}) + 2 πn = 2 π - arccos (\frac{3}{3}) + 2 πn

2 π - arccos (\frac{1}{3}) + 2 πn

= 2 π - arccos (\frac{3}{3}) + 2 πn

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

共通因数を約分する:

2 π - arccos (\frac{3}{3}) + 2 πn

= x

簡素化

\frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn} : \frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn}

\frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

2 π - arccos (\frac{1}{3}) + 2 πn = 2 π - arccos (\frac{3}{3}) + 2 πn

2 π - arccos (\frac{1}{3}) + 2 πn

= 2 π - arccos (\frac{3}{3}) + 2 πn

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

x = \frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

x = \frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

x = \frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

簡素化

\frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn} : \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

\frac{π}{2 π - arccos ( \frac{3}{3} ) + 2 πn}

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}}}{3}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{3 ^{1 - \frac{1}{2}}}

数を引く:

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

= \frac{1}{3 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{1}{3}

= \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

x = \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}; n \neq = \frac{- 2 π + arccos ( \frac{1}{3} )}{2 π}

x = \frac{π}{arccos ( \frac{1}{3} ) + 2 πn}, x = \frac{π}{2 π - arccos ( \frac{1}{3} ) + 2 πn}

10進法形式で解を証明する

x = \frac{π}{0.95531 \dots + 2 πn}, x = \frac{π}{2 π - 0.95531 \dots + 2 πn}

グラフ

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