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人気のある三角関数 >

cos(1/(3x))= 1/3

解

cos (\frac{1}{3 x}) = \frac{1}{3}

解

x = \frac{1}{3 ( 1.23095 \dots + 2 πn )}, x = \frac{1}{3 ( 2 π - 1.23095 \dots + 2 πn )}

+1

度

x = 0^{\circ} + 2.54168 \dots^{\circ} n, x = 0^{\circ} + 1.68486 \dots^{\circ} n

解答ステップ

cos (\frac{1}{3 x}) = \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (\frac{1}{3 x}) = \frac{1}{3}

以下の一般解

cos (\frac{1}{3 x}) = \frac{1}{3}

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

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

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

解く

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

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

以下で両辺を乗じる:

x

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

以下で両辺を乗じる:

x

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

簡素化

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

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

辺を交換する

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

因数

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

arccos (\frac{1}{3}) x + 2 πn x

共通項をくくり出す

x

= x (arccos (\frac{1}{3}) + 2 πn)

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

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) = \frac{1}{3}

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

共通因数を約分する:

arccos (\frac{1}{3}) + 2 πn

= x

簡素化

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

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

分数の規則を適用する:

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

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

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

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

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

解く

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

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

以下で両辺を乗じる:

x

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

以下で両辺を乗じる:

x

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

簡素化

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

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

辺を交換する

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

因数

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

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

共通項をくくり出す

x

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

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

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) = \frac{1}{3}

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

共通因数を約分する:

2 π - arccos (\frac{1}{3}) + 2 πn

= x

簡素化

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

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

分数の規則を適用する:

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

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

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

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

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

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

10進法形式で解を証明する

x = \frac{1}{3 ( 1.23095 \dots + 2 πn )}, x = \frac{1}{3 ( 2 π - 1.23095 \dots + 2 πn )}

グラフ

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