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受欢迎的三角函数 >

csc^{(2)}(x)-cot(x)=2,0<,x<360

解答

csc^{(2)} (x) - cot (x) = 2, 0 <, x < 360

解答

x \in R 无解

求解步骤

csc^{(2)} (x) - cot (x) = 2, 0, x < 360

两边减去

2

csc^{2} (x) - cot (x) - 2 = 0

使用三角恒等式改写

- 2 - cot (x) + csc^{2} (x)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

= - 2 - cot (x) + 1 + cot^{2} (x)

化简

- 2 - cot (x) + 1 + cot^{2} (x) : cot^{2} (x) - cot (x) - 1

- 2 - cot (x) + 1 + cot^{2} (x)

对同类项分组

= - cot (x) + cot^{2} (x) - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

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

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

u_{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

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

将解分隔开

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

u = \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}

u = \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:

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

u = cot (x)

代回

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

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

cot (x) = \frac{1 + 5}{2}, 0 :

无解

cot (x) = \frac{1 + 5}{2}, 0

使用反三角函数性质

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

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

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1 + 5}{2}) + πn

x = arccot (\frac{1 + 5}{2}) + πn

在

0

范围内的解

无解

cot (x) = \frac{1 - 5}{2}, 0 :

无解

cot (x) = \frac{1 - 5}{2}, 0

使用反三角函数性质

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

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

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (\frac{1 - 5}{2}) + πn

x = arccot (\frac{1 - 5}{2}) + πn

在

0

范围内的解

无解

合并所有解

x \in R 无解

作图

流行的例子

20=27sin(x)-1.5cos(x)

20 = 27 sin (x) - 1.5 cos (x)

sin(x)= 1/(sec(x))

sin (x) = \frac{1}{sec ( x )}

cos (12 x) = 0

sin (α) = - \frac{3}{5}

solvefor x,y= 1/pi arctan(x/s)+1/2

so l v e f or x, y = \frac{1}{π} arctan (\frac{x}{s}) + \frac{1}{2}