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受欢迎的三角函数 >

sin^{sin(x)}(x)=2

解答

sin^{s i n (x)} (x) = 2

解答

x \in R 无解

求解步骤

sin^{s i n (x)} (x) = 2

用替代法求解

sin^{s i n (x)} (x) = 2

令

: sin (x) = u

u^{u} = 2

u^{u} = 2 : u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u^{u} = 2

将

u^{u} = 2

调整为朗伯形式:

u e^{- \frac{l n ( 2 )}{u}} = 1

u^{u} = 2

x e^{x} = a

是朗伯形式的方程

对方程式两边

\frac{1}{u}

次方

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

化简

(u^{u})^{\frac{1}{u}} : u

(u^{u})^{\frac{1}{u}}

使用指数法则:

(a^{b})^{c} = a^{b c},

假定

a \geq 0

= u^{u \frac{1}{u}}

u \frac{1}{u} = 1

u \frac{1}{u}

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

= u

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

在两边乘以

2^{- \frac{1}{u}}

u \cdot 2^{- \frac{1}{u}} = 2^{\frac{1}{u}} \cdot 2^{- \frac{1}{u}}

化简

2^{\frac{1}{u}} \cdot 2^{- \frac{1}{u}} : 1

2^{\frac{1}{u}} \cdot 2^{- \frac{1}{u}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{\frac{1}{u}} \cdot 2^{- \frac{1}{u}} = 2^{\frac{1}{u} - \frac{1}{u}}

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

同类项相加:

1 \cdot \frac{1}{u} - 1 \cdot \frac{1}{u} = 0

= 2^{0}

使用法则

a^{0} = 1, a \neq = 0

= 1

u \cdot 2^{- \frac{1}{u}} = 1

使用指数运算法则

u \cdot 2^{- \frac{1}{u}} = 1

将转换为以

e

为底数的幂:

u e^{l n (2) (- \frac{1}{u})} = 1

使用指数法则:

a = b^{l o g_{b} (a)}

2^{- \frac{1}{u}} = (e^{l n (2)})^{- \frac{1}{u}}

u (e^{l n (2)})^{- \frac{1}{u}} = 1

使用指数法则:

(a^{b})^{c} = a^{b c}

(e^{l n (2)})^{- \frac{1}{u}} = e^{l n (2) (- \frac{1}{u})}

u e^{l n (2) (- \frac{1}{u})} = 1

u e^{l n (2) (- \frac{1}{u})} = 1

化简

u e^{- \frac{l n ( 2 )}{u}} = 1

u e^{- \frac{l n ( 2 )}{u}} = 1

用

\frac{ln ( 2 )}{u} = v

和

u = \frac{ln ( 2 )}{v}

改写方程式

(\frac{ln ( 2 )}{v}) e^{- v} = 1

改写

(\frac{ln ( 2 )}{v}) e^{- v} = 1

为朗伯形式:

v e^{v} = ln (2)

(\frac{ln ( 2 )}{v}) e^{- v} = 1

x e^{x} = a

是朗伯形式的方程

在两边乘以

v

\frac{ln ( 2 )}{v} e^{- v} v = 1 \cdot v

化简

ln (2) e^{- v} = v

在两边乘以

e^{v}

ln (2) e^{- v} e^{v} = v e^{v}

化简

ln (2) e^{- v} e^{v} : ln (2)

ln (2) e^{- v} e^{v}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

e^{- v} e^{v} = e^{- v + v}

= ln (2) e^{- v + v}

同类项相加:

- v + v = 0

= ln (2) e^{0}

使用法则

a^{0} = 1, a \neq = 0

= 1 \cdot ln (2)

乘以:

ln (2) \cdot 1 = ln (2)

= ln (2)

ln (2) = v e^{v}

交换两边

v e^{v} = ln (2)

解

v e^{v} = ln (2) : v = W_{0} (ln (2))

v e^{v} = ln (2)

x e^{x} = a

（

a \geq 0

）的解为朗伯

W

函数的主分支:

x = W_{0} (a)

v = W_{0} (ln (2))

验证解:

v = W_{0} (ln (2))

真

将它们代入

(\frac{ln ( 2 )}{v}) e^{- v} = 1

检验解是否符合

去除与方程不符的解。

代入

v = W_{0} (ln (2)) :

真

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))} = 1

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))} = \frac{e ^{- W_{0} (l n (2))} ln ( 2 )}{W _{0} ( ln ( 2 ) )}

(\frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}) e^{- W_{0} (l n (2))}

去除括号:

(a) = a

= \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )} e^{- W_{0} (l n (2))}

分式相乘:

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

= \frac{ln ( 2 ) e ^{- W_{0} (l n (2))}}{W _{0} ( ln ( 2 ) )}

\frac{e ^{- W_{0} (l n (2))} ln ( 2 )}{W _{0} ( ln ( 2 ) )} = 1

真

解是

v = W_{0} (ln (2))

代回

v = \frac{ln ( 2 )}{u}

，求解

u

解

\frac{ln ( 2 )}{u} = W_{0} (ln (2)) : u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

\frac{ln ( 2 )}{u} = W_{0} (ln (2))

在两边乘以

u

\frac{ln ( 2 )}{u} = W_{0} (ln (2))

在两边乘以

u

\frac{ln ( 2 )}{u} u = W_{0} (ln (2)) u

化简

ln (2) = W_{0} (ln (2)) u

ln (2) = W_{0} (ln (2)) u

交换两边

W_{0} (ln (2)) u = ln (2)

两边除以

W_{0} (ln (2))

W_{0} (ln (2)) u = ln (2)

两边除以

W_{0} (ln (2))

\frac{W _{0} ( ln ( 2 ) ) u}{W _{0} ( ln ( 2 ) )} = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

化简

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{ln ( 2 )}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

u = sin (x)

代回

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )} :

无解

sin (x) = \frac{ln ( 2 )}{W _{0} ( ln ( 2 ) )}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x \in R 无解

作图

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