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受欢迎的三角函数 >

csc(X)-cot(X)=5

解答

csc (X) - cot (X) = 5

解答

X = 2.74680 \dots + 2 πn

+1

度数

X = 157.38013 \dots^{\circ} + 36 0^{\circ} n

求解步骤

csc (X) - cot (X) = 5

两边减去

5

csc (X) - cot (X) - 5 = 0

用 sin, cos 表示

\frac{1}{sin ( X )} - \frac{cos ( X )}{sin ( X )} - 5 = 0

化简

\frac{1}{sin ( X )} - \frac{cos ( X )}{sin ( X )} - 5 : \frac{1 - cos ( X ) - 5 sin ( X )}{sin ( X )}

\frac{1}{sin ( X )} - \frac{cos ( X )}{sin ( X )} - 5

合并分式

\frac{1}{sin ( X )} - \frac{cos ( X )}{sin ( X )} : \frac{1 - cos ( X )}{sin ( X )}

使用法则

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

= \frac{1 - cos ( X )}{sin ( X )}

= \frac{- cos ( X ) + 1}{sin ( X )} - 5

将项转换为分式:

5 = \frac{5 sin ( X )}{sin ( X )}

= \frac{1 - cos ( X )}{sin ( X )} - \frac{5 sin ( X )}{sin ( X )}

因为分母相等，所以合并分式:

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

= \frac{1 - cos ( X ) - 5 sin ( X )}{sin ( X )}

\frac{1 - cos ( X ) - 5 sin ( X )}{sin ( X )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - cos (X) - 5 sin (X) = 0

两边加上

5 sin (X)

1 - cos (X) = 5 sin (X)

两边进行平方

(1 - cos (X))^{2} = (5 sin (X))^{2}

两边减去

(5 sin (X))^{2}

(1 - cos (X))^{2} - 25 sin^{2} (X) = 0

使用三角恒等式改写

(1 - cos (X))^{2} - 25 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)

= (1 - cos (X))^{2} - 25 (1 - cos^{2} (X))

化简

(1 - cos (X))^{2} - 25 (1 - cos^{2} (X)) : 26 cos^{2} (X) - 2 cos (X) - 24

(1 - cos (X))^{2} - 25 (1 - cos^{2} (X))

(1 - cos (X))^{2} : 1 - 2 cos (X) + cos^{2} (X)

使用完全平方公式:

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

a = 1, b = cos (X)

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

化简

1^{2} - 2 \cdot 1 \cdot cos (X) + cos^{2} (X) : 1 - 2 cos (X) + cos^{2} (X)

1^{2} - 2 \cdot 1 \cdot cos (X) + cos^{2} (X)

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

= 1 - 2 cos (X) + cos^{2} (X)

= 1 - 2 cos (X) + cos^{2} (X)

= 1 - 2 cos (X) + cos^{2} (X) - 25 (1 - cos^{2} (X))

乘开

- 25 (1 - cos^{2} (X)) : - 25 + 25 cos^{2} (X)

- 25 (1 - cos^{2} (X))

使用分配律:

a (b - c) = ab - a c

a = - 25, b = 1, c = cos^{2} (X)

= - 25 \cdot 1 - (- 25) cos^{2} (X)

使用加减运算法则

- (- a) = a

= - 25 \cdot 1 + 25 cos^{2} (X)

数字相乘:

25 \cdot 1 = 25

= - 25 + 25 cos^{2} (X)

= 1 - 2 cos (X) + cos^{2} (X) - 25 + 25 cos^{2} (X)

化简

1 - 2 cos (X) + cos^{2} (X) - 25 + 25 cos^{2} (X) : 26 cos^{2} (X) - 2 cos (X) - 24

1 - 2 cos (X) + cos^{2} (X) - 25 + 25 cos^{2} (X)

对同类项分组

= - 2 cos (X) + cos^{2} (X) + 25 cos^{2} (X) + 1 - 25

同类项相加:

cos^{2} (X) + 25 cos^{2} (X) = 26 cos^{2} (X)

= - 2 cos (X) + 26 cos^{2} (X) + 1 - 25

数字相加/相减:

1 - 25 = - 24

= 26 cos^{2} (X) - 2 cos (X) - 24

= 26 cos^{2} (X) - 2 cos (X) - 24

= 26 cos^{2} (X) - 2 cos (X) - 24

- 24 + 26 cos^{2} (X) - 2 cos (X) = 0

用替代法求解

- 24 + 26 cos^{2} (X) - 2 cos (X) = 0

令

: cos (X) = u

- 24 + 26 u^{2} - 2 u = 0

- 24 + 26 u^{2} - 2 u = 0 : u = 1, u = - \frac{12}{13}

- 24 + 26 u^{2} - 2 u = 0

改写成标准形式

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

26 u^{2} - 2 u - 24 = 0

使用求根公式求解

26 u^{2} - 2 u - 24 = 0

二次方程求根公式:

若

a = 26, b = - 2, c = - 24

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 26 ( - 24 )}{2 \cdot 26}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 26 ( - 24 )}{2 \cdot 26}

(- 2)^{2} - 4 \cdot 26 (- 24) = 50

(- 2)^{2} - 4 \cdot 26 (- 24)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 26 \cdot 24

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 26 \cdot 24

数字相乘:

4 \cdot 26 \cdot 24 = 2496

= 2^{2} + 2496

2^{2} = 4

= 4 + 2496

数字相加:

4 + 2496 = 2500

= 2500

因式分解数字:

2500 = 5 0^{2}

= 5 0^{2}

使用根式运算法则:

n a^{n} = a

5 0^{2} = 50

= 50

u_{1, 2} = \frac{- ( - 2 ) \pm 50}{2 \cdot 26}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 50}{2 \cdot 26}, u_{2} = \frac{- ( - 2 ) - 50}{2 \cdot 26}

u = \frac{- ( - 2 ) + 50}{2 \cdot 26} : 1

\frac{- ( - 2 ) + 50}{2 \cdot 26}

使用法则

- (- a) = a

= \frac{2 + 50}{2 \cdot 26}

数字相加:

2 + 50 = 52

= \frac{52}{2 \cdot 26}

数字相乘:

2 \cdot 26 = 52

= \frac{52}{52}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 50}{2 \cdot 26} : - \frac{12}{13}

\frac{- ( - 2 ) - 50}{2 \cdot 26}

使用法则

- (- a) = a

= \frac{2 - 50}{2 \cdot 26}

数字相减:

2 - 50 = - 48

= \frac{- 48}{2 \cdot 26}

数字相乘:

2 \cdot 26 = 52

= \frac{- 48}{52}

使用分式法则:

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

= - \frac{48}{52}

约分:

4

= - \frac{12}{13}

二次方程组的解是:

u = 1, u = - \frac{12}{13}

u = cos (X)

代回

cos (X) = 1, cos (X) = - \frac{12}{13}

cos (X) = 1, cos (X) = - \frac{12}{13}

cos (X) = 1 : X = 2 πn

cos (X) = 1

cos (X) = 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

X = 0 + 2 πn

X = 0 + 2 πn

解

X = 0 + 2 πn : X = 2 πn

X = 0 + 2 πn

0 + 2 πn = 2 πn

X = 2 πn

X = 2 πn

cos (X) = - \frac{12}{13} : X = arccos (- \frac{12}{13}) + 2 πn, X = - arccos (- \frac{12}{13}) + 2 πn

cos (X) = - \frac{12}{13}

使用反三角函数性质

cos (X) = - \frac{12}{13}

cos (X) = - \frac{12}{13}

的通解

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

X = arccos (- \frac{12}{13}) + 2 πn, X = - arccos (- \frac{12}{13}) + 2 πn

X = arccos (- \frac{12}{13}) + 2 πn, X = - arccos (- \frac{12}{13}) + 2 πn

合并所有解

X = 2 πn, X = arccos (- \frac{12}{13}) + 2 πn, X = - arccos (- \frac{12}{13}) + 2 πn

将解代入原方程进行验证

将它们代入

csc (X) - cot (X) = 5

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

假

2 πn

代入

n = 1

2 π 1

对于

csc (X) - cot (X) = 5

代入

X = 2 π 1

csc (2 π 1) - cot (2 π 1) = 5

未定义

\Rightarrow 假

检验

arccos (- \frac{12}{13}) + 2 πn

的解:

真

arccos (- \frac{12}{13}) + 2 πn

代入

n = 1

arccos (- \frac{12}{13}) + 2 π 1

对于

csc (X) - cot (X) = 5

代入

X = arccos (- \frac{12}{13}) + 2 π 1

csc (arccos (- \frac{12}{13}) + 2 π 1) - cot (arccos (- \frac{12}{13}) + 2 π 1) = 5

整理后得

5 = 5

\Rightarrow 真

检验

- arccos (- \frac{12}{13}) + 2 πn

的解:

假

- arccos (- \frac{12}{13}) + 2 πn

代入

n = 1

- arccos (- \frac{12}{13}) + 2 π 1

对于

csc (X) - cot (X) = 5

代入

X = - arccos (- \frac{12}{13}) + 2 π 1

csc (- arccos (- \frac{12}{13}) + 2 π 1) - cot (- arccos (- \frac{12}{13}) + 2 π 1) = 5

整理后得

- 5 = 5

\Rightarrow 假

X = arccos (- \frac{12}{13}) + 2 πn

以小数形式表示解

X = 2.74680 \dots + 2 πn

作图

流行的例子

-6cos^2(θ)=-cos(θ)-2

- 6 cos^{2} (θ) = - cos (θ) - 2

cos(2x)-3cos(-x)+2=0

cos (2 x) - 3 cos (- x) + 2 = 0

tan(x)+cot(x)= 5/2

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

sin(2x)+sqrt(2)cos(x)=0

sin (2 x) + 2 cos (x) = 0

csc(2x)=-sqrt(2)

csc (2 x) = - 2