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受欢迎的三角函数 >

3sinh(x)+cosh(x)=-2

解答

3 sinh (x) + cosh (x) = - 2

解答

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

+1

度数

x = - 57.58526 \dots^{\circ}

求解步骤

3 sinh (x) + cosh (x) = - 2

使用三角恒等式改写

3 sinh (x) + cosh (x) = - 2

使用双曲函数恒等式:

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

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

使用双曲函数恒等式:

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

3 \cdot \frac{e ^{x} - e ^{- x}}{2} + \frac{e ^{x} + e ^{- x}}{2} = - 2

3 \cdot \frac{e ^{x} - e ^{- x}}{2} + \frac{e ^{x} + e ^{- x}}{2} = - 2

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

3 \cdot \frac{e ^{x} - e ^{- x}}{2} + \frac{e ^{x} + e ^{- x}}{2} = - 2

在两边乘以

2

3 \cdot \frac{e ^{x} - e ^{- x}}{2} \cdot 2 + \frac{e ^{x} + e ^{- x}}{2} \cdot 2 = - 2 \cdot 2

化简

3 (e^{x} - e^{- x}) + e^{x} + e^{- x} = - 4

使用指数运算法则

3 (e^{x} - e^{- x}) + e^{x} + e^{- x} = - 4

使用指数法则:

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

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

3 (e^{x} - (e^{x})^{- 1}) + e^{x} + (e^{x})^{- 1} = - 4

3 (e^{x} - (e^{x})^{- 1}) + e^{x} + (e^{x})^{- 1} = - 4

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

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

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

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

展开

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

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

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

乘开

3 u (u - \frac{1}{u}) : 3 u^{2} - 3

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

3 uu - 3 \cdot \frac{1}{u} u : 3 u^{2} - 3

3 uu - 3 \cdot \frac{1}{u} u

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}

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

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

分式相乘:

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

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

约分:

u

= 1 \cdot 3

数字相乘:

1 \cdot 3 = 3

= 3

= 3 u^{2} - 3

= 3 u^{2} - 3

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

化简

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

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

对同类项分组

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

同类项相加:

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

= 4 u^{2} - 3 + 1

数字相加/相减:

- 3 + 1 = - 2

= 4 u^{2} - 2

= 4 u^{2} - 2

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

解

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

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

将

4 u

para o lado esquerdo

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

两边加上

4 u

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

化简

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

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

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 4, c = - 2

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

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

4^{2} - 4 \cdot 4 (- 2) = 43

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 4^{2} + 32

4^{2} = 16

= 16 + 32

数字相加:

16 + 32 = 48

= 48

48

质因数分解:

2^{4} \cdot 3

48

48

除以

248 = 24 \cdot 2

= 2 \cdot 24

24

除以

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{4} \cdot 3

= 2^{4} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{4}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

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

= 2^{2} 3

整理后得

= 43

u_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 4 + 4 3}{2 \cdot 4}, u_{2} = \frac{- 4 - 4 3}{2 \cdot 4}

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

\frac{- 4 + 4 3}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 + 4 3}{8}

分解

- 4 + 43 : 4 (- 1 + 3)

- 4 + 43

改写为

= - 4 \cdot 1 + 43

因式分解出通项

4

= 4 (- 1 + 3)

= \frac{4 ( - 1 + 3 )}{8}

约分:

4

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

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

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 - 4 3}{8}

分解

- 4 - 43 : - 4 (1 + 3)

- 4 - 43

改写为

= - 4 \cdot 1 - 43

因式分解出通项

4

= - 4 (1 + 3)

= - \frac{4 ( 1 + 3 )}{8}

约分:

4

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

二次方程组的解是:

u = \frac{- 1 + 3}{2}, u = - \frac{1 + 3}{2}

u = \frac{- 1 + 3}{2}, u = - \frac{1 + 3}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = \frac{- 1 + 3}{2}, u = - \frac{1 + 3}{2}

代回

u = e^{x}

，求解

x

解

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

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

使用指数运算法则

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

若

f (x) = g (x)

，则

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

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

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

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

解

e^{x} = - \frac{1 + 3}{2} : x \in R

无解

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

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

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

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

作图

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