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受欢迎的三角函数 >

5cosh(x)-3sinh(x)=5

解答

5 cosh (x) - 3 sinh (x) = 5

解答

x = 2 ln (2), x = 0

+1

度数

x = 79.42881 \dots^{\circ}, x = 0^{\circ}

求解步骤

5 cosh (x) - 3 sinh (x) = 5

使用三角恒等式改写

5 cosh (x) - 3 sinh (x) = 5

使用双曲函数恒等式:

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

5 cosh (x) - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

使用双曲函数恒等式:

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

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5 : x = 2 ln (2), x = 0

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

使用指数运算法则

5 \cdot \frac{e ^{x} + e ^{- x}}{2} - 3 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5

使用指数法则:

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

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

5 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} - 3 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

5 \cdot \frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2} - 3 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

用

e^{x} = u

改写方程式

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

解

5 \cdot \frac{u + u ^{- 1}}{2} - 3 \cdot \frac{u - u ^{- 1}}{2} = 5 : u = 4, u = 1

5 \cdot \frac{u + u ^{- 1}}{2} - 3 \cdot \frac{u - u ^{- 1}}{2} = 5

整理后得

\frac{5 ( u ^{2} + 1 )}{2 u} - \frac{3 ( u ^{2} - 1 )}{2 u} = 5

在两边乘以

2 u

\frac{5 ( u ^{2} + 1 )}{2 u} - \frac{3 ( u ^{2} - 1 )}{2 u} = 5

在两边乘以

2 u

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

化简

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

化简

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u : 5 (u^{2} + 1)

\frac{5 ( u ^{2} + 1 )}{2 u} \cdot 2 u

分式相乘:

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

= \frac{5 ( u ^{2} + 1 ) \cdot 2 u}{2 u}

约分:

2

= \frac{5 ( u ^{2} + 1 ) u}{u}

约分:

u

= 5 (u^{2} + 1)

化简

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

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

分式相乘:

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

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

约分:

2

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

约分:

u

= - 3 (u^{2} - 1)

化简

5 \cdot 2 u : 10 u

5 \cdot 2 u

数字相乘:

5 \cdot 2 = 10

= 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

解

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u : u = 4, u = 1

5 (u^{2} + 1) - 3 (u^{2} - 1) = 10 u

展开

5 (u^{2} + 1) - 3 (u^{2} - 1) : 2 u^{2} + 8

5 (u^{2} + 1) - 3 (u^{2} - 1)

乘开

5 (u^{2} + 1) : 5 u^{2} + 5

5 (u^{2} + 1)

使用分配律:

a (b + c) = ab + a c

a = 5, b = u^{2}, c = 1

= 5 u^{2} + 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 5 u^{2} + 5

= 5 u^{2} + 5 - 3 (u^{2} - 1)

乘开

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

- 3 (u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = - 3, b = u^{2}, c = 1

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

使用加减运算法则

- (- a) = a

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

数字相乘:

3 \cdot 1 = 3

= - 3 u^{2} + 3

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

化简

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

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

对同类项分组

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

同类项相加:

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

= 2 u^{2} + 5 + 3

数字相加:

5 + 3 = 8

= 2 u^{2} + 8

= 2 u^{2} + 8

2 u^{2} + 8 = 10 u

将

10 u

para o lado esquerdo

2 u^{2} + 8 = 10 u

两边减去

10 u

2 u^{2} + 8 - 10 u = 10 u - 10 u

化简

2 u^{2} + 8 - 10 u = 0

2 u^{2} + 8 - 10 u = 0

改写成标准形式

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

2 u^{2} - 10 u + 8 = 0

使用求根公式求解

2 u^{2} - 10 u + 8 = 0

二次方程求根公式:

若

a = 2, b = - 10, c = 8

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 \cdot 8 = 6

(- 10)^{2} - 4 \cdot 2 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 2 \cdot 8

数字相乘:

4 \cdot 2 \cdot 8 = 64

= 1 0^{2} - 64

1 0^{2} = 100

= 100 - 64

数字相减:

100 - 64 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 10 ) \pm 6}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 6}{2 \cdot 2}, u_{2} = \frac{- ( - 10 ) - 6}{2 \cdot 2}

u = \frac{- ( - 10 ) + 6}{2 \cdot 2} : 4

\frac{- ( - 10 ) + 6}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{10 + 6}{2 \cdot 2}

数字相加:

10 + 6 = 16

= \frac{16}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{16}{4}

数字相除:

\frac{16}{4} = 4

= 4

u = \frac{- ( - 10 ) - 6}{2 \cdot 2} : 1

\frac{- ( - 10 ) - 6}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{10 - 6}{2 \cdot 2}

数字相减:

10 - 6 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 4, u = 1

u = 4, u = 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = 4, u = 1

代回

u = e^{x}

，求解

x

解

e^{x} = 4 : x = 2 ln (2)

e^{x} = 4

使用指数运算法则

e^{x} = 4

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (4)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (4)

化简

ln (4) : 2 ln (2)

ln (4)

改写

4

为幂形式:

4 = 2^{2}

= ln (2^{2})

使用对数计算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= 2 ln (2)

x = 2 ln (2)

x = 2 ln (2)

解

e^{x} = 1 : x = 0

e^{x} = 1

使用指数运算法则

e^{x} = 1

若

f (x) = g (x)

，则

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

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

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

化简

ln (1) : 0

ln (1)

使用对数计算法则:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

x = 2 ln (2), x = 0

x = 2 ln (2), x = 0

作图

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