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 >

e^{cos(x)}sin^2(x)-e^{cos(x)}cos(x)=0

解答

e^{c o s (x)} sin^{2} (x) - e^{c o s (x)} cos (x) = 0

解答

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度数

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

e^{c o s (x)} sin^{2} (x) - e^{c o s (x)} cos (x) = 0

使用三角恒等式改写

- cos (x) e^{c o s (x)} + sin^{2} (x) e^{c o s (x)}

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - cos (x) e^{c o s (x)} + (1 - cos^{2} (x)) e^{c o s (x)}

(1 - cos^{2} (x)) e^{c o s (x)} - cos (x) e^{c o s (x)} = 0

用替代法求解

(1 - cos^{2} (x)) e^{c o s (x)} - cos (x) e^{c o s (x)} = 0

令

: cos (x) = u

(1 - u^{2}) e^{u} - u e^{u} = 0

(1 - u^{2}) e^{u} - u e^{u} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

(1 - u^{2}) e^{u} - u e^{u} = 0

因式分解

e^{u} (- u^{2} + 1 - u) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

e^{u} = 0 or - u^{2} + 1 - u = 0

解

e^{u} = 0 : u \in R

无解

e^{u} = 0

a^{f (u)}

对于

u

不能为零或负值

\in R

u \in R 无解

解

- u^{2} + 1 - u = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

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

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

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

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

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

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

u = cos (x)

代回

cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{5 - 1}{2}

cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{5 - 1}{2}

cos (x) = - \frac{1 + 5}{2} :

无解

cos (x) = - \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{5 - 1}{2} : x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

cos (x) = \frac{5 - 1}{2}

使用反三角函数性质

cos (x) = \frac{5 - 1}{2}

cos (x) = \frac{5 - 1}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

合并所有解

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

以小数形式表示解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

2cot(x)+tan(x)=-2

2 cot (x) + tan (x) = - 2

sin(x)-sqrt(3)=0

sin (x) - 3 = 0

2 tan (4 x) = 1

cos(θ)=(5713634.83)/((2236)(3733))

cos (θ) = \frac{5713634.83}{( 2236 ) ( 3733 )}

sin(x)= 3/(sqrt(58))

sin (x) = \frac{3}{58}