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 >

sin^2(x)+cos^4(x)=2

解答

sin^{2} (x) + cos^{4} (x) = 2

解答

x \in R 无解

求解步骤

sin^{2} (x) + cos^{4} (x) = 2

两边减去

2

sin^{2} (x) + cos^{4} (x) - 2 = 0

使用三角恒等式改写

- 2 + cos^{4} (x) + sin^{2} (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)

= - 2 + cos^{4} (x) + 1 - cos^{2} (x)

化简

- 2 + cos^{4} (x) + 1 - cos^{2} (x) : cos^{4} (x) - cos^{2} (x) - 1

- 2 + cos^{4} (x) + 1 - cos^{2} (x)

对同类项分组

= cos^{4} (x) - cos^{2} (x) - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

= cos^{4} (x) - cos^{2} (x) - 1

= cos^{4} (x) - cos^{2} (x) - 1

- 1 - cos^{2} (x) + cos^{4} (x) = 0

用替代法求解

- 1 - cos^{2} (x) + cos^{4} (x) = 0

令

: cos (x) = u

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

- 1 - u^{2} + u^{4} = 0 : u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

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

改写成标准形式

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - v - 1 = 0

解

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

v^{2} - v - 1 = 0

使用求根公式求解

v^{2} - v - 1 = 0

二次方程求根公式:

若

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

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

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

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

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

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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

u^{2} = \frac{1 + 5}{2}

对于

x^{2} = f (a)

解为

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

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

解

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

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

对于

x^{2} = f (a)

解为

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

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

解为

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

u = cos (x)

代回

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

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

无解

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

- 1 \leq cos (x) \leq 1

无解

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

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

因为方程对以下值无定义:

arccos \frac{1 - 5}{2} + 2 πn, - arccos \frac{1 - 5}{2} + 2 πn, arccos - \frac{1 - 5}{2} + 2 πn, - arccos - \frac{1 - 5}{2} + 2 πn

x \in R 无解

作图

流行的例子

(1+sin^2(x))=(4cos(x))^2

(1 + sin^{2} (x)) = (4 cos (x))^{2}

cos (x) = 53

sin^2(a)=(1-cos(a))/2

sin^{2} (a) = \frac{1 - cos ( a )}{2}

cos(x)=(-11)/(12)

cos (x) = \frac{- 11}{12}

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

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