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人気のある三角関数 >

solvefor x,Rf=e^{cos(2x)}

解

解く

x, R f = e^{c o s (2 x)}

解

x = \frac{arccos ( ln ( R f ) )}{2} + πn, x = - \frac{arccos ( ln ( R f ) )}{2} + πn

解答ステップ

R f = e^{c o s (2 x)}

辺を交換する

e^{c o s (2 x)} = R f

指数の規則を適用する

e^{c o s (2 x)} = R f

f (x) = g (x)

ならば,

ln (f (x)) = ln (g (x))

ln (e^{c o s (2 x)}) = ln (R f)

対数の規則を適用する:

ln (e^{a}) = a

ln (e^{c o s (2 x)}) = cos (2 x)

cos (2 x) = ln (R f)

cos (2 x) = ln (R f)

三角関数の逆数プロパティを適用する

cos (2 x) = ln (R f)

以下の一般解

cos (2 x) = ln (R f)

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

2 x = arccos (ln (R f)) + 2 πn, 2 x = - arccos (ln (R f)) + 2 πn

2 x = arccos (ln (R f)) + 2 πn, 2 x = - arccos (ln (R f)) + 2 πn

解く

2 x = arccos (ln (R f)) + 2 πn : x = \frac{arccos ( ln ( R f ) )}{2} + πn

2 x = arccos (ln (R f)) + 2 πn

以下で両辺を割る

2

2 x = arccos (ln (R f)) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{arccos ( ln ( R f ) )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{arccos ( ln ( R f ) )}{2} + πn

x = \frac{arccos ( ln ( R f ) )}{2} + πn

解く

2 x = - arccos (ln (R f)) + 2 πn : x = - \frac{arccos ( ln ( R f ) )}{2} + πn

2 x = - arccos (ln (R f)) + 2 πn

以下で両辺を割る

2

2 x = - arccos (ln (R f)) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = - \frac{arccos ( ln ( R f ) )}{2} + \frac{2 πn}{2}

簡素化

x = - \frac{arccos ( ln ( R f ) )}{2} + πn

x = - \frac{arccos ( ln ( R f ) )}{2} + πn

x = \frac{arccos ( ln ( R f ) )}{2} + πn, x = - \frac{arccos ( ln ( R f ) )}{2} + πn

グラフ

人気の例

sin(2θ)= 12/13

sin (2 θ) = \frac{12}{13}

sin^2(θ)+sin(θ)=2

sin^{2} (θ) + sin (θ) = 2

7sin^2(θ)-3sin(θ)=4

7 sin^{2} (θ) - 3 sin (θ) = 4

- 5 - 4 tan (θ) = - 1

sin (θ) = - \frac{7}{10}