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受欢迎的三角函数 >

1+7sinh(x)=4cosh^2(x)

解答

1 + 7 sinh (x) = 4 cosh^{2} (x)

解答

x = ln (2), x = ln (1 + 2)

+1

度数

x = 39.71440 \dots^{\circ}, x = 50.49898 \dots^{\circ}

求解步骤

1 + 7 sinh (x) = 4 cosh^{2} (x)

使用三角恒等式改写

1 + 7 sinh (x) = 4 cosh^{2} (x)

使用双曲函数恒等式:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 cosh^{2} (x)

使用双曲函数恒等式:

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

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2} : x = ln (2), x = ln (1 + 2)

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

使用指数运算法则

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

使用指数法则:

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

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

1 + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 4 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 4 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2}

用

e^{x} = u

改写方程式

1 + 7 \cdot \frac{u - ( u ) ^{- 1}}{2} = 4 (\frac{u + ( u ) ^{- 1}}{2})^{2}

解

1 + 7 \cdot \frac{u - u ^{- 1}}{2} = 4 (\frac{u + u ^{- 1}}{2})^{2} : u = 2, u = - \frac{1}{2}, u = 1 + 2, u = 1 - 2

1 + 7 \cdot \frac{u - u ^{- 1}}{2} = 4 (\frac{u + u ^{- 1}}{2})^{2}

整理后得

1 + \frac{7 ( u ^{2} - 1 )}{2 u} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}}

乘以最小公倍数

1 + \frac{7 ( u ^{2} - 1 )}{2 u} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}}

找到

2 u, u^{2}

的最小公倍数:

2 u^{2}

2 u, u^{2}

最小公倍数 (LCM)

计算出由出现在

2 u

或

u^{2}

中的因子组成的表达式

= 2 u^{2}

乘以最小公倍数=

2 u^{2}

1 \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2}

化简

1 \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2}

化简

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

1 \cdot 2 u^{2}

数字相乘:

1 \cdot 2 = 2

= 2 u^{2}

化简

\frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} : 7 u (u^{2} - 1)

\frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2}

分式相乘:

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

= \frac{7 ( u ^{2} - 1 ) \cdot 2 u ^{2}}{2 u}

约分:

2

= \frac{7 ( u ^{2} - 1 ) u ^{2}}{u}

约分:

u

= 7 u (u^{2} - 1)

化简

\frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2} : 2 (u^{2} + 1)^{2}

\frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2}

分式相乘:

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

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

约分:

u^{2}

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

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

解

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2} : u = 2, u = - \frac{1}{2}, u = 1 + 2, u = 1 - 2

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

展开

2 u^{2} + 7 u (u^{2} - 1) : 2 u^{2} + 7 u^{3} - 7 u

2 u^{2} + 7 u (u^{2} - 1)

乘开

7 u (u^{2} - 1) : 7 u^{3} - 7 u

7 u (u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = 7 u, b = u^{2}, c = 1

= 7 u u^{2} - 7 u \cdot 1

= 7 u^{2} u - 7 \cdot 1 \cdot u

化简

7 u^{2} u - 7 \cdot 1 \cdot u : 7 u^{3} - 7 u

7 u^{2} u - 7 \cdot 1 \cdot u

7 u^{2} u = 7 u^{3}

7 u^{2} u

使用指数法则:

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

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

= 7 u^{2 + 1}

数字相加:

2 + 1 = 3

= 7 u^{3}

7 \cdot 1 \cdot u = 7 u

7 \cdot 1 \cdot u

数字相乘:

7 \cdot 1 = 7

= 7 u

= 7 u^{3} - 7 u

= 7 u^{3} - 7 u

= 2 u^{2} + 7 u^{3} - 7 u

展开

2 (u^{2} + 1)^{2} : 2 u^{4} + 4 u^{2} + 2

2 (u^{2} + 1)^{2}

(u^{2} + 1)^{2} = u^{4} + 2 u^{2} + 1

(u^{2} + 1)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

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

化简

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

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

使用法则

1^{a} = 1

1^{2} = 1

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

(u^{2})^{2} = u^{4}

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

= 2 (u^{4} + 2 u^{2} + 1)

打开括号

= 2 u^{4} + 2 \cdot 2 u^{2} + 2 \cdot 1

化简

2 u^{4} + 2 \cdot 2 u^{2} + 2 \cdot 1 : 2 u^{4} + 4 u^{2} + 2

2 u^{4} + 2 \cdot 2 u^{2} + 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 2 u^{4} + 4 u^{2} + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{4} + 4 u^{2} + 2

= 2 u^{4} + 4 u^{2} + 2

2 u^{2} + 7 u^{3} - 7 u = 2 u^{4} + 4 u^{2} + 2

交换两边

2 u^{4} + 4 u^{2} + 2 = 2 u^{2} + 7 u^{3} - 7 u

将

7 u

para o lado esquerdo

2 u^{4} + 4 u^{2} + 2 = 2 u^{2} + 7 u^{3} - 7 u

两边加上

7 u

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3} - 7 u + 7 u

化简

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

将

7 u^{3}

para o lado esquerdo

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

两边减去

7 u^{3}

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2} + 7 u^{3} - 7 u^{3}

化简

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

将

2 u^{2}

para o lado esquerdo

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

两边减去

2 u^{2}

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} - 2 u^{2} = 2 u^{2} - 2 u^{2}

化简

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 = 0

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 = 0

因式分解

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 : (u - 2) (2 u + 1) (u^{2} - 2 u - 1)

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

使用有理根定理

a_{0} = 2, a_{n} = 2

a_{0}

的除数

: 1, 2, a_{n}

的除数

: 1, 2

因此，检验以下有理数:

\pm \frac{1 , 2}{1 , 2}

\frac{2}{1}

是表达式的根，所以因式分解

u - 2

= (u - 2) \frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} - 3 u^{2} - 4 u - 1

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

对

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

做除法:

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

将分子

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

与除数

u - 2

的首项系数相除：

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

商 = 2 u^{3}

将

u - 2

乘以

2 u^{3} : 2 u^{4} - 4 u^{3}

将

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

减去

2 u^{4} - 4 u^{3}

得到新的余数

余数 = - 3 u^{3} + 2 u^{2} + 7 u + 2

因此

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

= 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

对

\frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

做除法:

\frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

将分子

- 3 u^{3} + 2 u^{2} + 7 u + 2

与除数

u - 2

的首项系数相除：

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

商 = - 3 u^{2}

将

u - 2

乘以

- 3 u^{2} : - 3 u^{3} + 6 u^{2}

将

- 3 u^{3} + 2 u^{2} + 7 u + 2

减去

- 3 u^{3} + 6 u^{2}

得到新的余数

余数 = - 4 u^{2} + 7 u + 2

因此

\frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

= 2 u^{3} - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

对

\frac{- 4 u ^{2} + 7 u + 2}{u - 2}

做除法:

\frac{- 4 u ^{2} + 7 u + 2}{u - 2} = - 4 u + \frac{- u + 2}{u - 2}

将分子

- 4 u^{2} + 7 u + 2

与除数

u - 2

的首项系数相除：

\frac{- 4 u ^{2}}{u} = - 4 u

商 = - 4 u

将

u - 2

乘以

- 4 u : - 4 u^{2} + 8 u

将

- 4 u^{2} + 7 u + 2

减去

- 4 u^{2} + 8 u

得到新的余数

余数 = - u + 2

因此

\frac{- 4 u ^{2} + 7 u + 2}{u - 2} = - 4 u + \frac{- u + 2}{u - 2}

= 2 u^{3} - 3 u^{2} - 4 u + \frac{- u + 2}{u - 2}

对

\frac{- u + 2}{u - 2}

做除法:

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

将分子

- u + 2

与除数

u - 2

的首项系数相除：

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

商 = - 1

将

u - 2

乘以

- 1 : - u + 2

将

- u + 2

减去

- u + 2

得到新的余数

余数 = 0

因此

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

= 2 u^{3} - 3 u^{2} - 4 u - 1

= 2 u^{3} - 3 u^{2} - 4 u - 1

分解

2 u^{3} - 3 u^{2} - 4 u - 1 : (2 u + 1) (u^{2} - 2 u - 1)

2 u^{3} - 3 u^{2} - 4 u - 1

使用有理根定理

a_{0} = 1, a_{n} = 2

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2

因此，检验以下有理数:

\pm \frac{1}{1 , 2}

- \frac{1}{2}

是表达式的根，所以因式分解

2 u + 1

= (2 u + 1) \frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1}

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} - 2 u - 1

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1}

对

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1}

做除法:

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} + \frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

将分子

2 u^{3} - 3 u^{2} - 4 u - 1

与除数

2 u + 1

的首项系数相除：

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

商 = u^{2}

将

2 u + 1

乘以

u^{2} : 2 u^{3} + u^{2}

将

2 u^{3} - 3 u^{2} - 4 u - 1

减去

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

得到新的余数

余数 = - 4 u^{2} - 4 u - 1

因此

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} + \frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

= u^{2} + \frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

对

\frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

做除法:

\frac{- 4 u ^{2} - 4 u - 1}{2 u + 1} = - 2 u + \frac{- 2 u - 1}{2 u + 1}

将分子

- 4 u^{2} - 4 u - 1

与除数

2 u + 1

的首项系数相除：

\frac{- 4 u ^{2}}{2 u} = - 2 u

商 = - 2 u

将

2 u + 1

乘以

- 2 u : - 4 u^{2} - 2 u

将

- 4 u^{2} - 4 u - 1

减去

- 4 u^{2} - 2 u

得到新的余数

余数 = - 2 u - 1

因此

\frac{- 4 u ^{2} - 4 u - 1}{2 u + 1} = - 2 u + \frac{- 2 u - 1}{2 u + 1}

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

对

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

做除法:

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

将分子

- 2 u - 1

与除数

2 u + 1

的首项系数相除：

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

商 = - 1

将

2 u + 1

乘以

- 1 : - 2 u - 1

将

- 2 u - 1

减去

- 2 u - 1

得到新的余数

余数 = 0

因此

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

= u^{2} - 2 u - 1

= u^{2} - 2 u - 1

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

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

(u - 2) (2 u + 1) (u^{2} - 2 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 2 = 0 or 2 u + 1 = 0 or u^{2} - 2 u - 1 = 0

解

u - 2 = 0 : u = 2

u - 2 = 0

将

2

到右边

u - 2 = 0

两边加上

2

u - 2 + 2 = 0 + 2

化简

u = 2

u = 2

解

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

u = - \frac{1}{2}

u = - \frac{1}{2}

解

u^{2} - 2 u - 1 = 0 : u = 1 + 2, u = 1 - 2

u^{2} - 2 u - 1 = 0

使用求根公式求解

u^{2} - 2 u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 2, c = - 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

(- 2)^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

u_{1, 2} = \frac{- ( - 2 ) \pm 2 2}{2 \cdot 1}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

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

数字相除:

\frac{2}{2} = 1

= 1 + 2

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

分解

2 - 22 : 2 (1 - 2)

2 - 22

改写为

= 2 \cdot 1 - 22

因式分解出通项

2

= 2 (1 - 2)

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

数字相除:

\frac{2}{2} = 1

= 1 - 2

二次方程组的解是:

u = 1 + 2, u = 1 - 2

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

1 + 7 \frac{u - u ^{- 1}}{2}

的分母，令其等于零

u = 0

取

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

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

u = 2, u = - \frac{1}{2}, u = 1 + 2, u = 1 - 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} = - \frac{1}{2} : x \in R

无解

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

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

e^{x} = 1 + 2 : x = ln (1 + 2)

e^{x} = 1 + 2

使用指数运算法则

e^{x} = 1 + 2

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (1 + 2)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1 + 2)

x = ln (1 + 2)

解

e^{x} = 1 - 2 : x \in R

无解

e^{x} = 1 - 2

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = ln (2), x = ln (1 + 2)

x = ln (2), x = ln (1 + 2)

作图

流行的例子

sin^2(x)-1+2cos(2x)-cos^2(x)=0

sin^{2} (x) - 1 + 2 cos (2 x) - cos^{2} (x) = 0

cos^2(x)+6cos(x)+5=0

cos^{2} (x) + 6 cos (x) + 5 = 0

sin (t) = - 0.9397

tan^2(x)=2sec^2(x)-3

tan^{2} (x) = 2 sec^{2} (x) - 3

sinh(x)+4=4cosh(x)

sinh (x) + 4 = 4 cosh (x)