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受欢迎的三角函数 >

tanh(x)= 1/2

解答

tanh (x) = \frac{1}{2}

解答

x = \frac{1}{2} ln (3)

+1

度数

x = 31.47292 \dots^{\circ}

求解步骤

tanh (x) = \frac{1}{2}

使用三角恒等式改写

tanh (x) = \frac{1}{2}

使用双曲函数恒等式:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{1}{2}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{1}{2}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{1}{2} : x = \frac{1}{2} ln (3)

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{1}{2}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

(e^{x} - e^{- x}) \cdot 2 = (e^{x} + e^{- x}) \cdot 1

化简

(e^{x} - e^{- x}) \cdot 2 = e^{x} + e^{- x}

使用指数运算法则

(e^{x} - e^{- x}) \cdot 2 = e^{x} + e^{- x}

使用指数法则:

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

e^{- x} = (e^{x})^{- 1}

(e^{x} - (e^{x})^{- 1}) \cdot 2 = e^{x} + (e^{x})^{- 1}

(e^{x} - (e^{x})^{- 1}) \cdot 2 = e^{x} + (e^{x})^{- 1}

用

e^{x} = u

改写方程式

(u - (u)^{- 1}) \cdot 2 = u + (u)^{- 1}

解

(u - u^{- 1}) \cdot 2 = u + u^{- 1} : u = 3, u = - 3

(u - u^{- 1}) \cdot 2 = u + u^{- 1}

整理后得

(u - \frac{1}{u}) \cdot 2 = u + \frac{1}{u}

化简

(u - \frac{1}{u}) \cdot 2 : 2 (u - \frac{1}{u})

(u - \frac{1}{u}) \cdot 2

使用交换律:

(u - \frac{1}{u}) \cdot 2 = 2 (u - \frac{1}{u})

2 (u - \frac{1}{u})

2 (u - \frac{1}{u}) = u + \frac{1}{u}

在两边乘以

u

2 (u - \frac{1}{u}) = u + \frac{1}{u}

在两边乘以

u

2 (u - \frac{1}{u}) u = uu + \frac{1}{u} u

化简

2 (u - \frac{1}{u}) u = uu + \frac{1}{u} u

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

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

约分:

u

= 1

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

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

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

展开

2 (u - \frac{1}{u}) u : 2 u^{2} - 2

2 (u - \frac{1}{u}) u

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

使用分配律:

a (b - c) = ab - a c

a = 2 u, b = u, c = \frac{1}{u}

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

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

化简

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

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

2 uu = 2 u^{2}

2 uu

使用指数法则:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= 2 u^{2}

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

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

分式相乘:

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

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

约分:

u

= 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= 2

= 2 u^{2} - 2

= 2 u^{2} - 2

2 u^{2} - 2 = u^{2} + 1

将

2

到右边

2 u^{2} - 2 = u^{2} + 1

两边加上

2

2 u^{2} - 2 + 2 = u^{2} + 1 + 2

化简

2 u^{2} = u^{2} + 3

2 u^{2} = u^{2} + 3

解

2 u^{2} = u^{2} + 3 : u = 3, u = - 3

2 u^{2} = u^{2} + 3

将

u^{2}

para o lado esquerdo

2 u^{2} = u^{2} + 3

两边减去

u^{2}

2 u^{2} - u^{2} = u^{2} + 3 - u^{2}

化简

u^{2} = 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

u = 3, u = - 3

验证解

找到无定义的点（奇点）:

u = 0

取

(u - u^{- 1}) 2

的分母，令其等于零

u = 0

取

u + u^{- 1}

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 = 3, u = - 3

u = 3, u = - 3

代回

u = e^{x}

，求解

x

解

e^{x} = 3 : x = \frac{1}{2} ln (3)

e^{x} = 3

使用指数运算法则

e^{x} = 3

使用指数法则:

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

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

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

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (3^{\frac{1}{2}})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (3^{\frac{1}{2}})

使用对数计算法则:

ln (x^{a}) = a \cdot ln (x)

ln (3^{\frac{1}{2}}) = \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3)

解

e^{x} = - 3 : x \in R

无解

e^{x} = - 3

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = \frac{1}{2} ln (3)

验证解:

x = \frac{1}{2} ln (3)

真

将它们代入

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{1}{2}

检验解是否符合

去除与方程不符的解。

代入

x = \frac{1}{2} ln (3) :

真

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}} = \frac{1}{2}

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}} = \frac{1}{2}

\frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{e ^{\frac{1}{2} l n (3)} + e ^{- \frac{1}{2} l n (3)}}

e^{\frac{1}{2} l n (3)} = 3

e^{\frac{1}{2} l n (3)}

使用指数法则:

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

= e^{l n (3)}

使用对数计算法则:

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

e^{l n (3)} = 3

= 3

e^{- \frac{1}{2} l n (3)} = 3^{- \frac{1}{2}}

e^{- \frac{1}{2} l n (3)}

使用指数法则:

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

= (e^{l n (3)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (3)} = 3

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

= \frac{e ^{\frac{1}{2} l n (3)} - e ^{- \frac{1}{2} l n (3)}}{3 + 3 ^{- \frac{1}{2}}}

e^{\frac{1}{2} l n (3)} = 3

e^{\frac{1}{2} l n (3)}

使用指数法则:

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

= e^{l n (3)}

使用对数计算法则:

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

e^{l n (3)} = 3

= 3

e^{- \frac{1}{2} l n (3)} = 3^{- \frac{1}{2}}

e^{- \frac{1}{2} l n (3)}

使用指数法则:

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

= (e^{l n (3)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (3)} = 3

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

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

化简

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

使用指数法则:

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

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

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

使用指数法则:

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

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

= \frac{3 - \frac{1}{3}}{3 + \frac{1}{3}}

化简

3 + \frac{1}{3} : \frac{4}{3}

3 + \frac{1}{3}

将项转换为分式:

3 = \frac{3 3}{3}

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

因为分母相等，所以合并分式:

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

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

33 + 1 = 4

33 + 1

使用根式运算法则:

a a = a

33 = 3

= 3 + 1

数字相加:

3 + 1 = 4

= 4

= \frac{4}{3}

= \frac{3 - \frac{1}{3}}{\frac{4}{3}}

化简

3 - \frac{1}{3} : \frac{2}{3}

3 - \frac{1}{3}

将项转换为分式:

3 = \frac{3 3}{3}

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

因为分母相等，所以合并分式:

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

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

33 - 1 = 2

33 - 1

使用根式运算法则:

a a = a

33 = 3

= 3 - 1

数字相减:

3 - 1 = 2

= 2

= \frac{2}{3}

= \frac{\frac{2}{3}}{\frac{4}{3}}

分式相除:

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

= \frac{2 3}{3 \cdot 4}

约分:

3

= \frac{2}{4}

约分:

2

= \frac{1}{2}

= \frac{1}{2}

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

真

解是

x = \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3)

作图

流行的例子

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2 cos (x) - 2 sin (x) = 2

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solvefor x,f=arctan(x/(sqrt(1-x^2)))

so l v e f or x, f = arctan (\frac{x}{1 - x ^{2}})

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