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 >

1=sech^2(x)

解答

1 = sech^{2} (x)

解答

x = 0

+1

度数

x = 0^{\circ}

求解步骤

1 = sech^{2} (x)

交换两边

sech^{2} (x) = 1

使用三角恒等式改写

sech^{2} (x) = 1

使用双曲函数恒等式:

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

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

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

(\frac{2}{e ^{x} + e ^{- x}})^{2} = 1 : x = 0

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

使用指数运算法则

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

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

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

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

在两边乘以

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

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

化简

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

化简

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

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

分式相乘:

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

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

约分:

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

= 4 u^{2}

化简

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

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

乘以:

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

= (u^{2} + 1)^{2}

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

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

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

解

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

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

展开

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

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

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

(u^{2})^{2} = u^{4}

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

交换两边

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

将

4 u^{2}

para o lado esquerdo

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

两边减去

4 u^{2}

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

化简

u^{4} - 2 u^{2} + 1 = 0

u^{4} - 2 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 2 v + 1 = 0

解

v^{2} - 2 v + 1 = 0 : v = 1

v^{2} - 2 v + 1 = 0

使用求根公式求解

v^{2} - 2 v + 1 = 0

二次方程求根公式:

若

a = 1, b = - 2, c = 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

(- 2)^{2} - 4 \cdot 1 \cdot 1

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数字相减:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

\frac{- ( - 2 )}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

v = 1

二次方程组的解是:

v = 1

v = 1

代回

v = u^{2}

，求解

u

解

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

解为

u = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = 1, u = - 1

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} = - 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 = 0

x = 0

作图

流行的例子

solvefor t,fw=2+cos(10pit)

so l v e f or t, f w = 2 + cos (10 π t)

tan(θ)-sec(θ)=sqrt(3)

tan (θ) - sec (θ) = 3

cos (t) = \frac{24}{25}

sec^2(t)+2sec(t)=0

sec^{2} (t) + 2 sec (t) = 0

sin(270+x)-cos(180-x)=-sin(x)

sin (27 0^{\circ} + x) - cos (18 0^{\circ} - x) = - sin (x)