zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tanh(x)= 3/5

解答

tanh (x) = \frac{3}{5}

解答

x = ln (2)

+1

度数

x = 39.71440 \dots^{\circ}

求解步骤

tanh (x) = \frac{3}{5}

使用三角恒等式改写

tanh (x) = \frac{3}{5}

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{3}{5}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5} : x = ln (2)

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

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 5 = (e^{x} + e^{- x}) \cdot 3

使用指数运算法则

(e^{x} - e^{- x}) \cdot 5 = (e^{x} + e^{- x}) \cdot 3

使用指数法则:

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

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

(e^{x} - (e^{x})^{- 1}) \cdot 5 = (e^{x} + (e^{x})^{- 1}) \cdot 3

(e^{x} - (e^{x})^{- 1}) \cdot 5 = (e^{x} + (e^{x})^{- 1}) \cdot 3

用

e^{x} = u

改写方程式

(u - (u)^{- 1}) \cdot 5 = (u + (u)^{- 1}) \cdot 3

解

(u - u^{- 1}) \cdot 5 = (u + u^{- 1}) \cdot 3 : u = 2, u = - 2

(u - u^{- 1}) \cdot 5 = (u + u^{- 1}) \cdot 3

整理后得

(u - \frac{1}{u}) \cdot 5 = (u + \frac{1}{u}) \cdot 3

化简

(u - \frac{1}{u}) \cdot 5 = (u + \frac{1}{u}) \cdot 3

化简

(u - \frac{1}{u}) \cdot 5 : 5 (u - \frac{1}{u})

(u - \frac{1}{u}) \cdot 5

使用交换律:

(u - \frac{1}{u}) \cdot 5 = 5 (u - \frac{1}{u})

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

化简

(u + \frac{1}{u}) \cdot 3 : 3 (u + \frac{1}{u})

(u + \frac{1}{u}) \cdot 3

使用交换律:

(u + \frac{1}{u}) \cdot 3 = 3 (u + \frac{1}{u})

3 (u + \frac{1}{u})

5 (u - \frac{1}{u}) = 3 (u + \frac{1}{u})

5 (u - \frac{1}{u}) = 3 (u + \frac{1}{u})

展开

5 (u - \frac{1}{u}) : 5 u - \frac{5}{u}

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

使用分配律:

a (b - c) = ab - a c

a = 5, b = u, c = \frac{1}{u}

= 5 u - 5 \cdot \frac{1}{u}

5 \cdot \frac{1}{u} = \frac{5}{u}

5 \cdot \frac{1}{u}

分式相乘:

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

= \frac{1 \cdot 5}{u}

数字相乘:

1 \cdot 5 = 5

= \frac{5}{u}

= 5 u - \frac{5}{u}

展开

3 (u + \frac{1}{u}) : 3 u + \frac{3}{u}

3 (u + \frac{1}{u})

使用分配律:

a (b + c) = ab + a c

a = 3, b = u, c = \frac{1}{u}

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

3 \cdot \frac{1}{u} = \frac{3}{u}

3 \cdot \frac{1}{u}

分式相乘:

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

= \frac{1 \cdot 3}{u}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{u}

= 3 u + \frac{3}{u}

5 u - \frac{5}{u} = 3 u + \frac{3}{u}

在两边乘以

u

5 u - \frac{5}{u} = 3 u + \frac{3}{u}

在两边乘以

u

5 uu - \frac{5}{u} u = 3 uu + \frac{3}{u} u

化简

5 uu - \frac{5}{u} u = 3 uu + \frac{3}{u} u

化简

5 uu : 5 u^{2}

5 uu

使用指数法则:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

数字相加:

1 + 1 = 2

= 5 u^{2}

化简

- \frac{5}{u} u : - 5

- \frac{5}{u} u

分式相乘:

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

= - \frac{5 u}{u}

约分:

u

= - 5

化简

3 uu : 3 u^{2}

3 uu

使用指数法则:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

数字相加:

1 + 1 = 2

= 3 u^{2}

化简

\frac{3}{u} u : 3

\frac{3}{u} u

分式相乘:

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

= \frac{3 u}{u}

约分:

u

= 3

5 u^{2} - 5 = 3 u^{2} + 3

5 u^{2} - 5 = 3 u^{2} + 3

5 u^{2} - 5 = 3 u^{2} + 3

解

5 u^{2} - 5 = 3 u^{2} + 3 : u = 2, u = - 2

5 u^{2} - 5 = 3 u^{2} + 3

将

5

到右边

5 u^{2} - 5 = 3 u^{2} + 3

两边加上

5

5 u^{2} - 5 + 5 = 3 u^{2} + 3 + 5

化简

5 u^{2} = 3 u^{2} + 8

5 u^{2} = 3 u^{2} + 8

将

3 u^{2}

para o lado esquerdo

5 u^{2} = 3 u^{2} + 8

两边减去

3 u^{2}

5 u^{2} - 3 u^{2} = 3 u^{2} + 8 - 3 u^{2}

化简

2 u^{2} = 8

2 u^{2} = 8

两边除以

2

2 u^{2} = 8

两边除以

2

\frac{2 u ^{2}}{2} = \frac{8}{2}

化简

u^{2} = 4

u^{2} = 4

对于

x^{2} = f (a)

解为

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

u = 4, u = - 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

- 4 = - 2

- 4

因式分解数字:

4 = 2^{2}

= - 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = - 2

= - 2

u = 2, u = - 2

u = 2, u = - 2

验证解

找到无定义的点（奇点）:

u = 0

取

(u - u^{- 1}) 5

的分母，令其等于零

u = 0

取

(u + u^{- 1}) 3

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 = 2, u = - 2

u = 2, u = - 2

代回

u = e^{x}

，求解

x

解

e^{x} = 2 : x = ln (2)

e^{x} = 2

使用指数运算法则

e^{x} = 2

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (2)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2)

x = ln (2)

解

e^{x} = - 2 : x \in R

无解

e^{x} = - 2

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = ln (2)

验证解:

x = ln (2)

真

将它们代入

\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = \frac{3}{5}

检验解是否符合

去除与方程不符的解。

代入

x = ln (2) :

真

\frac{e ^{l n (2)} - e ^{- l n (2)}}{e ^{l n (2)} + e ^{- l n (2)}} = \frac{3}{5}

\frac{e ^{l n (2)} - e ^{- l n (2)}}{e ^{l n (2)} + e ^{- l n (2)}} = \frac{3}{5}

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

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

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

= 2

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

e^{- l n (2)}

使用指数法则:

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

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

使用对数计算法则:

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

e^{l n (2)} = 2

= 2^{- 1}

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

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

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

= 2

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

e^{- l n (2)}

使用指数法则:

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

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

使用对数计算法则:

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

e^{l n (2)} = 2

= 2^{- 1}

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

化简

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

化简

2 + \frac{1}{2} : \frac{5}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}

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

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{2 \cdot 2 + 1}{2}

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

数字相乘:

2 \cdot 2 = 4

= 4 + 1

数字相加:

4 + 1 = 5

= 5

= \frac{5}{2}

= \frac{2 - \frac{1}{2}}{\frac{5}{2}}

化简

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

2 - \frac{1}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}

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

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{2 \cdot 2 - 1}{2}

2 \cdot 2 - 1 = 3

2 \cdot 2 - 1

数字相乘:

2 \cdot 2 = 4

= 4 - 1

数字相减:

4 - 1 = 3

= 3

= \frac{3}{2}

= \frac{\frac{3}{2}}{\frac{5}{2}}

分式相除:

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

= \frac{3 \cdot 2}{2 \cdot 5}

约分:

2

= \frac{3}{5}

= \frac{3}{5}

\frac{3}{5} = \frac{3}{5}

真

解是

x = ln (2)

x = ln (2)

作图

流行的例子

csc(θ)+2.402=0

csc (θ) + 2.402 = 0

tan^{2} (β) = 1

6sin(C)+sqrt(8)=0

6 sin (C) + 8 = 0

cos(2t)=-sin(t)

cos (2 t) = - sin (t)

sin^2(x)-5cos(x)-5=0

sin^{2} (x) - 5 cos (x) - 5 = 0