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受欢迎的三角函数 >

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

解答

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

解答

x = - 2 ln (2)

+1

度数

x = - 79.42881 \dots^{\circ}

求解步骤

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

使用三角恒等式改写

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

使用双曲函数恒等式:

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

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

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

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

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

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

使用指数运算法则

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

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

2 u

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

在两边乘以

2 u

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

化简

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

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

\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

- 3 \cdot 2 u : - 6 u

- 3 \cdot 2 u

数字相乘:

3 \cdot 2 = 6

= - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

解

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u : u = \frac{1}{4}, u = - 1

3 (u^{2} + 1) + 5 (u^{2} - 1) = - 6 u

展开

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

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

数字相乘:

3 \cdot 1 = 3

= 3 u^{2} + 3

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

= 8 u^{2} + 3 - 5

数字相加/相减:

3 - 5 = - 2

= 8 u^{2} - 2

= 8 u^{2} - 2

8 u^{2} - 2 = - 6 u

将

6 u

para o lado esquerdo

8 u^{2} - 2 = - 6 u

两边加上

6 u

8 u^{2} - 2 + 6 u = - 6 u + 6 u

化简

8 u^{2} - 2 + 6 u = 0

8 u^{2} - 2 + 6 u = 0

改写成标准形式

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

8 u^{2} + 6 u - 2 = 0

使用求根公式求解

8 u^{2} + 6 u - 2 = 0

二次方程求根公式:

若

a = 8, b = 6, c = - 2

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 2 )}{2 \cdot 8}

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

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

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 8 \cdot 2

数字相乘:

4 \cdot 8 \cdot 2 = 64

= 6^{2} + 64

6^{2} = 36

= 36 + 64

数字相加:

36 + 64 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 6 \pm 10}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 6 + 10}{2 \cdot 8}, u_{2} = \frac{- 6 - 10}{2 \cdot 8}

u = \frac{- 6 + 10}{2 \cdot 8} : \frac{1}{4}

\frac{- 6 + 10}{2 \cdot 8}

数字相加/相减:

- 6 + 10 = 4

= \frac{4}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{4}{16}

约分:

4

= \frac{1}{4}

u = \frac{- 6 - 10}{2 \cdot 8} : - 1

\frac{- 6 - 10}{2 \cdot 8}

数字相减:

- 6 - 10 = - 16

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 16}{16}

使用分式法则:

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

= - \frac{16}{16}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{1}{4}, u = - 1

u = \frac{1}{4}, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = \frac{1}{4}, u = - 1

代回

u = e^{x}

，求解

x

解

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

e^{x} = \frac{1}{4}

使用指数运算法则

e^{x} = \frac{1}{4}

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (\frac{1}{4})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{1}{4})

化简

ln (\frac{1}{4}) : - 2 ln (2)

ln (\frac{1}{4})

使用对数计算法则:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - 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 \in R

无解

e^{x} = - 1

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = - 2 ln (2)

x = - 2 ln (2)

作图

流行的例子

6cos^2(x)-5cos(x)+1=0

6 cos^{2} (x) - 5 cos (x) + 1 = 0

6 cos (x) = 5

6cos(2x)=6cos^2(x)-5

6 cos (2 x) = 6 cos^{2} (x) - 5

(-4cos(x)-6sin(x))^2-20sin^2(x)=16

(- 4 cos (x) - 6 sin (x))^{2} - 20 sin^{2} (x) = 16

sin (u) = \frac{5}{13}