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受欢迎的三角函数 >

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

解答

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

解答

x = 0, x = ln (\frac{5}{3})

+1

度数

x = 0^{\circ}, x = 29.26815 \dots^{\circ}

求解步骤

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

使用三角恒等式改写

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

使用双曲函数恒等式:

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

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

使用双曲函数恒等式:

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

\frac{e ^{x} - e ^{- x}}{2} + 4 = 4 \cdot \frac{e ^{x} + e ^{- x}}{2}

\frac{e ^{x} - e ^{- x}}{2} + 4 = 4 \cdot \frac{e ^{x} + e ^{- x}}{2}

\frac{e ^{x} - e ^{- x}}{2} + 4 = 4 \cdot \frac{e ^{x} + e ^{- x}}{2} : x = 0, x = ln (\frac{5}{3})

\frac{e ^{x} - e ^{- x}}{2} + 4 = 4 \cdot \frac{e ^{x} + e ^{- x}}{2}

在两边乘以

2

\frac{e ^{x} - e ^{- x}}{2} \cdot 2 + 4 \cdot 2 = 4 \cdot \frac{e ^{x} + e ^{- x}}{2} \cdot 2

化简

e^{x} - e^{- x} + 8 = 4 (e^{x} + e^{- x})

使用指数运算法则

e^{x} - e^{- x} + 8 = 4 (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} + 8 = 4 (e^{x} + (e^{x})^{- 1})

e^{x} - (e^{x})^{- 1} + 8 = 4 (e^{x} + (e^{x})^{- 1})

用

e^{x} = u

改写方程式

u - (u)^{- 1} + 8 = 4 (u + (u)^{- 1})

解

u - u^{- 1} + 8 = 4 (u + u^{- 1}) : u = 1, u = \frac{5}{3}

u - u^{- 1} + 8 = 4 (u + u^{- 1})

整理后得

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

在两边乘以

u

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

在两边乘以

u

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

化简

uu - \frac{1}{u} u + 8 u = 4 (u + \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

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

u^{2} - 1 + 8 u = 4 (u + \frac{1}{u}) u

展开

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

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

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

使用分配律:

a (b + c) = ab + a c

a = 4 u, b = u, c = \frac{1}{u}

= 4 uu + 4 u \frac{1}{u}

= 4 uu + 4 \cdot \frac{1}{u} u

化简

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

4 uu + 4 \cdot \frac{1}{u} u

4 uu = 4 u^{2}

4 uu

使用指数法则:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

数字相加:

1 + 1 = 2

= 4 u^{2}

4 \cdot \frac{1}{u} u = 4

4 \cdot \frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot 4 u}{u}

约分:

u

= 1 \cdot 4

数字相乘:

1 \cdot 4 = 4

= 4

= 4 u^{2} + 4

= 4 u^{2} + 4

u^{2} - 1 + 8 u = 4 u^{2} + 4

将

1

到右边

u^{2} - 1 + 8 u = 4 u^{2} + 4

两边加上

1

u^{2} - 1 + 8 u + 1 = 4 u^{2} + 4 + 1

化简

u^{2} + 8 u = 4 u^{2} + 5

u^{2} + 8 u = 4 u^{2} + 5

解

u^{2} + 8 u = 4 u^{2} + 5 : u = 1, u = \frac{5}{3}

u^{2} + 8 u = 4 u^{2} + 5

将

5

para o lado esquerdo

u^{2} + 8 u = 4 u^{2} + 5

两边减去

5

u^{2} + 8 u - 5 = 4 u^{2} + 5 - 5

化简

u^{2} + 8 u - 5 = 4 u^{2}

u^{2} + 8 u - 5 = 4 u^{2}

将

4 u^{2}

para o lado esquerdo

u^{2} + 8 u - 5 = 4 u^{2}

两边减去

4 u^{2}

u^{2} + 8 u - 5 - 4 u^{2} = 4 u^{2} - 4 u^{2}

化简

- 3 u^{2} + 8 u - 5 = 0

- 3 u^{2} + 8 u - 5 = 0

使用求根公式求解

- 3 u^{2} + 8 u - 5 = 0

二次方程求根公式:

若

a = - 3, b = 8, c = - 5

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

8^{2} - 4 (- 3) (- 5) = 2

8^{2} - 4 (- 3) (- 5)

使用法则

- (- a) = a

= 8^{2} - 4 \cdot 3 \cdot 5

数字相乘:

4 \cdot 3 \cdot 5 = 60

= 8^{2} - 60

8^{2} = 64

= 64 - 60

数字相减:

64 - 60 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- 8 \pm 2}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 8 + 2}{2 ( - 3 )}, u_{2} = \frac{- 8 - 2}{2 ( - 3 )}

u = \frac{- 8 + 2}{2 ( - 3 )} : 1

\frac{- 8 + 2}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 8 + 2}{- 2 \cdot 3}

数字相加/相减:

- 8 + 2 = - 6

= \frac{- 6}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{- 6}

使用分式法则:

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

= \frac{6}{6}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- 8 - 2}{2 ( - 3 )} : \frac{5}{3}

\frac{- 8 - 2}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 8 - 2}{- 2 \cdot 3}

数字相减:

- 8 - 2 = - 10

= \frac{- 10}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 10}{- 6}

使用分式法则:

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

= \frac{10}{6}

约分:

2

= \frac{5}{3}

二次方程组的解是:

u = 1, u = \frac{5}{3}

u = 1, u = \frac{5}{3}

验证解

找到无定义的点（奇点）:

u = 0

取

u - u^{- 1} + 8

的分母，令其等于零

u = 0

取

4 (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 = 1, u = \frac{5}{3}

u = 1, u = \frac{5}{3}

代回

u = e^{x}

，求解

x

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

解

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

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

使用指数运算法则

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

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (\frac{5}{3})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

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

x = 0, x = ln (\frac{5}{3})

x = 0, x = ln (\frac{5}{3})

作图

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