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受欢迎的三角函数 >

5sinh(2x)+3cosh(2x)=4

解答

5 sinh (2 x) + 3 cosh (2 x) = 4

解答

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

+1

度数

x = 5.39228 \dots^{\circ}

求解步骤

5 sinh (2 x) + 3 cosh (2 x) = 4

使用三角恒等式改写

5 sinh (2 x) + 3 cosh (2 x) = 4

使用双曲函数恒等式:

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

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 cosh (2 x) = 4

使用双曲函数恒等式:

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

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4 : x = \frac{1}{2} ln (\frac{1 + 2}{2})

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

使用指数运算法则

5 \cdot \frac{e ^{2 x} - e ^{- 2 x}}{2} + 3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 4

使用指数法则:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

2 u^{2}

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

在两边乘以

2 u^{2}

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2} + \frac{3 ( u ^{4} + 1 )}{2 u ^{2}} \cdot 2 u^{2} = 4 \cdot 2 u^{2}

化简

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2} + \frac{3 ( u ^{4} + 1 )}{2 u ^{2}} \cdot 2 u^{2} = 4 \cdot 2 u^{2}

化简

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2} : 5 (u^{4} - 1)

\frac{5 ( u ^{4} - 1 )}{2 u ^{2}} \cdot 2 u^{2}

分式相乘:

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

= \frac{5 ( u ^{4} - 1 ) \cdot 2 u ^{2}}{2 u ^{2}}

约分:

2

= \frac{5 ( u ^{4} - 1 ) u ^{2}}{u ^{2}}

约分:

u^{2}

= 5 (u^{4} - 1)

化简

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

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

分式相乘:

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

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

约分:

2

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

约分:

u^{2}

= 3 (u^{4} + 1)

化简

4 \cdot 2 u^{2} : 8 u^{2}

4 \cdot 2 u^{2}

数字相乘:

4 \cdot 2 = 8

= 8 u^{2}

5 (u^{4} - 1) + 3 (u^{4} + 1) = 8 u^{2}

5 (u^{4} - 1) + 3 (u^{4} + 1) = 8 u^{2}

5 (u^{4} - 1) + 3 (u^{4} + 1) = 8 u^{2}

解

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

5 (u^{4} - 1) + 3 (u^{4} + 1) = 8 u^{2}

展开

5 (u^{4} - 1) + 3 (u^{4} + 1) : 8 u^{4} - 2

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

乘开

5 (u^{4} - 1) : 5 u^{4} - 5

5 (u^{4} - 1)

使用分配律:

a (b - c) = ab - a c

a = 5, b = u^{4}, c = 1

= 5 u^{4} - 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 5 u^{4} - 5

= 5 u^{4} - 5 + 3 (u^{4} + 1)

乘开

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

3 (u^{4} + 1)

使用分配律:

a (b + c) = ab + a c

a = 3, b = u^{4}, c = 1

= 3 u^{4} + 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 3 u^{4} + 3

= 5 u^{4} - 5 + 3 u^{4} + 3

化简

5 u^{4} - 5 + 3 u^{4} + 3 : 8 u^{4} - 2

5 u^{4} - 5 + 3 u^{4} + 3

对同类项分组

= 5 u^{4} + 3 u^{4} - 5 + 3

同类项相加:

5 u^{4} + 3 u^{4} = 8 u^{4}

= 8 u^{4} - 5 + 3

数字相加/相减:

- 5 + 3 = - 2

= 8 u^{4} - 2

= 8 u^{4} - 2

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

将

8 u^{2}

para o lado esquerdo

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

两边减去

8 u^{2}

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

化简

8 u^{4} - 2 - 8 u^{2} = 0

8 u^{4} - 2 - 8 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

8 u^{4} - 8 u^{2} - 2 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

8 v^{2} - 8 v - 2 = 0

解

8 v^{2} - 8 v - 2 = 0 : v = \frac{1 + 2}{2}, v = \frac{1 - 2}{2}

8 v^{2} - 8 v - 2 = 0

使用求根公式求解

8 v^{2} - 8 v - 2 = 0

二次方程求根公式:

若

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

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

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

(- 8)^{2} - 4 \cdot 8 (- 2) = 82

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 8 \cdot 2

数字相乘:

4 \cdot 8 \cdot 2 = 64

= 8^{2} + 64

8^{2} = 64

= 64 + 64

数字相加:

64 + 64 = 128

= 128

128

质因数分解:

2^{7}

128

128

除以

2128 = 64 \cdot 2

= 2 \cdot 64

64

除以

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

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

= 2^{7}

= 2^{7}

使用指数法则:

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

= 2^{6} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{6}

使用根式运算法则:

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

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 2

整理后得

= 82

v_{1, 2} = \frac{- ( - 8 ) \pm 8 2}{2 \cdot 8}

将解分隔开

v_{1} = \frac{- ( - 8 ) + 8 2}{2 \cdot 8}, v_{2} = \frac{- ( - 8 ) - 8 2}{2 \cdot 8}

v = \frac{- ( - 8 ) + 8 2}{2 \cdot 8} : \frac{1 + 2}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 8 = 16

= \frac{8 + 8 2}{16}

分解

8 + 82 : 8 (1 + 2)

8 + 82

改写为

= 8 \cdot 1 + 82

因式分解出通项

8

= 8 (1 + 2)

= \frac{8 ( 1 + 2 )}{16}

约分:

8

= \frac{1 + 2}{2}

v = \frac{- ( - 8 ) - 8 2}{2 \cdot 8} : \frac{1 - 2}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 8 = 16

= \frac{8 - 8 2}{16}

分解

8 - 82 : 8 (1 - 2)

8 - 82

改写为

= 8 \cdot 1 - 82

因式分解出通项

8

= 8 (1 - 2)

= \frac{8 ( 1 - 2 )}{16}

约分:

8

= \frac{1 - 2}{2}

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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

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

对于

x^{2} = f (a)

解为

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

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

解

u^{2} = \frac{1 - 2}{2} : u \in R

无解

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

x^{2}

在

x

内不能为负

\in R

u \in R 无解

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

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

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

代回

u = e^{x}

，求解

x

解

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

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

使用指数运算法则

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

使用指数法则:

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

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

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

若

f (x) = g (x)

，则

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

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

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

使用对数计算法则:

ln (x^{a}) = a \cdot ln (x)

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

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

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

解

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

无解

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

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

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

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

作图

流行的例子

sin (x) = 9

csc (θ) = 3

4sec(θ)=-5-tan^2(θ)

4 sec (θ) = - 5 - tan^{2} (θ)

cos(2x)=1-3sin(x)

cos (2 x) = 1 - 3 sin (x)

3sin(2t)+4=1,-pi/2 <= t<= pi/2

3 sin (2 t) + 4 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}