zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

3cosh(2x)=5

解答

3 cosh (2 x) = 5

解答

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

+1

度数

x = 31.47292 \dots^{\circ}, x = - 31.47292 \dots^{\circ}

求解步骤

3 cosh (2 x) = 5

使用三角恒等式改写

3 cosh (2 x) = 5

使用双曲函数恒等式:

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

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

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

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

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

使用指数运算法则

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

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

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

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

解

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

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

在两边乘以

2

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

在两边乘以

2

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

化简

3 (u^{4} + 1) = 10 u^{2}

3 (u^{4} + 1) = 10 u^{2}

展开

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

3 u^{4} + 3 = 10 u^{2}

将

10 u^{2}

para o lado esquerdo

3 u^{4} + 3 = 10 u^{2}

两边减去

10 u^{2}

3 u^{4} + 3 - 10 u^{2} = 10 u^{2} - 10 u^{2}

化简

3 u^{4} + 3 - 10 u^{2} = 0

3 u^{4} + 3 - 10 u^{2} = 0

改写成标准形式

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

3 u^{4} - 10 u^{2} + 3 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

3 v^{2} - 10 v + 3 = 0

解

3 v^{2} - 10 v + 3 = 0 : v = 3, v = \frac{1}{3}

3 v^{2} - 10 v + 3 = 0

使用求根公式求解

3 v^{2} - 10 v + 3 = 0

二次方程求根公式:

若

a = 3, b = - 10, c = 3

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 3 = 8

(- 10)^{2} - 4 \cdot 3 \cdot 3

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 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

数字相减:

100 - 36 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 3}

将解分隔开

v_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 3}, v_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 3}

v = \frac{- ( - 10 ) + 8}{2 \cdot 3} : 3

\frac{- ( - 10 ) + 8}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{10 + 8}{2 \cdot 3}

数字相加:

10 + 8 = 18

= \frac{18}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{18}{6}

数字相除:

\frac{18}{6} = 3

= 3

v = \frac{- ( - 10 ) - 8}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 10 ) - 8}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{10 - 8}{2 \cdot 3}

数字相减:

10 - 8 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

二次方程组的解是:

v = 3, v = \frac{1}{3}

v = 3, v = \frac{1}{3}

代回

v = u^{2}

，求解

u

解

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

解

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

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

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

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

\frac{1}{3} = \frac{1}{3}

\frac{1}{3}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{3}

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

- \frac{1}{3}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{3}

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

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

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

代回

u = e^{x}

，求解

x

解

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

e^{x} = 3

使用指数运算法则

e^{x} = 3

使用指数法则:

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

3 = 3^{\frac{1}{2}}

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

若

f (x) = g (x)

，则

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

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

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

使用对数计算法则:

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

ln (3^{\frac{1}{2}}) = \frac{1}{2} ln (3)

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

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

解

e^{x} = - 3 : x \in R

无解

e^{x} = - 3

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

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

e^{x} = \frac{1}{3}

使用指数运算法则

e^{x} = \frac{1}{3}

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

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

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

使用指数法则:

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

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

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

若

f (x) = g (x)

，则

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

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

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

使用对数计算法则:

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

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

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

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

解

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

无解

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

使用指数运算法则

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

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

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

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

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

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

作图

流行的例子

tan(2θ)=(2tan(θ))/(1-tan(θ))

tan (2 θ) = \frac{2 tan ( θ )}{1 - tan ( θ )}

sqrt(2)cos(θ)+1=0

2 cos (θ) + 1 = 0

4sqrt(2)tan(x)-sqrt(2)=3sqrt(2)tan(x)

42 tan (x) - 2 = 32 tan (x)

7sin(2x)+12cos(x)=0

7 sin (2 x) + 12 cos (x) = 0

solvefor θ,cos(θ)=-cos(40)

so l v e f or θ, cos (θ) = - cos (4 0^{\circ})