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

解答

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

解答

x \in R 无解

求解步骤

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

两边减去

2

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

用替代法求解

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

令

: sin (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 = sin (x)

代回

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

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

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

无解

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

- 1 \leq sin (x) \leq 1

无解

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

无解

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

- 1 \leq sin (x) \leq 1

无解

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

合并所有解

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

因为方程对以下值无定义:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

x \in R 无解

作图

流行的例子

5tan(x)=15cot(x)

5 tan (x) = 15 cot (x)

solvefor x,arctan(x)=arctan(y)

so l v e f or x, arctan (x) = arctan (y)

cos(x+10)-cos(x+90)=1

cos (x + 1 0^{\circ}) - cos (x + 9 0^{\circ}) = 1

2cos(x)-2sqrt(3)*sin(x)=sqrt(8)

2 cos (x) - 23 \cdot sin (x) = 8

((cot(x)-sqrt(3)))/((2sin(x)+1))=0

\frac{( cot ( x ) - 3 )}{( 2 sin ( x ) + 1 )} = 0