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 >

cot(θ)+4csc(θ)=6

解答

cot (θ) + 4 csc (θ) = 6

解答

θ = 0.88277 \dots + 2 πn, θ = 2.58911 \dots + 2 πn

+1

度数

θ = 50.57910 \dots^{\circ} + 36 0^{\circ} n, θ = 148.34553 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cot (θ) + 4 csc (θ) = 6

两边减去

6

cot (θ) + 4 csc (θ) - 6 = 0

用 sin, cos 表示

\frac{cos ( θ )}{sin ( θ )} + 4 \cdot \frac{1}{sin ( θ )} - 6 = 0

化简

\frac{cos ( θ )}{sin ( θ )} + 4 \cdot \frac{1}{sin ( θ )} - 6 : \frac{cos ( θ ) + 4 - 6 sin ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} + 4 \cdot \frac{1}{sin ( θ )} - 6

4 \cdot \frac{1}{sin ( θ )} = \frac{4}{sin ( θ )}

4 \cdot \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot 4}{sin ( θ )}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )} + \frac{4}{sin ( θ )} - 6

合并分式

\frac{cos ( θ )}{sin ( θ )} + \frac{4}{sin ( θ )} : \frac{cos ( θ ) + 4}{sin ( θ )}

使用法则

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

= \frac{cos ( θ ) + 4}{sin ( θ )}

= \frac{cos ( θ ) + 4}{sin ( θ )} - 6

将项转换为分式:

6 = \frac{6 sin ( θ )}{sin ( θ )}

= \frac{cos ( θ ) + 4}{sin ( θ )} - \frac{6 sin ( θ )}{sin ( θ )}

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{cos ( θ ) + 4 - 6 sin ( θ )}{sin ( θ )}

\frac{cos ( θ ) + 4 - 6 sin ( θ )}{sin ( θ )} = 0

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

cos (θ) + 4 - 6 sin (θ) = 0

两边加上

6 sin (θ)

cos (θ) + 4 = 6 sin (θ)

两边进行平方

(cos (θ) + 4)^{2} = (6 sin (θ))^{2}

两边减去

(6 sin (θ))^{2}

(cos (θ) + 4)^{2} - 36 sin^{2} (θ) = 0

使用三角恒等式改写

(4 + cos (θ))^{2} - 36 sin^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= (4 + cos (θ))^{2} - 36 (1 - cos^{2} (θ))

化简

(4 + cos (θ))^{2} - 36 (1 - cos^{2} (θ)) : 37 cos^{2} (θ) + 8 cos (θ) - 20

(4 + cos (θ))^{2} - 36 (1 - cos^{2} (θ))

(4 + cos (θ))^{2} : 16 + 8 cos (θ) + cos^{2} (θ)

使用完全平方公式:

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

a = 4, b = cos (θ)

= 4^{2} + 2 \cdot 4 cos (θ) + cos^{2} (θ)

化简

4^{2} + 2 \cdot 4 cos (θ) + cos^{2} (θ) : 16 + 8 cos (θ) + cos^{2} (θ)

4^{2} + 2 \cdot 4 cos (θ) + cos^{2} (θ)

4^{2} = 16

= 16 + 2 \cdot 4 cos (θ) + cos^{2} (θ)

数字相乘:

2 \cdot 4 = 8

= 16 + 8 cos (θ) + cos^{2} (θ)

= 16 + 8 cos (θ) + cos^{2} (θ)

= 16 + 8 cos (θ) + cos^{2} (θ) - 36 (1 - cos^{2} (θ))

乘开

- 36 (1 - cos^{2} (θ)) : - 36 + 36 cos^{2} (θ)

- 36 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 36, b = 1, c = cos^{2} (θ)

= - 36 \cdot 1 - (- 36) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 36 \cdot 1 + 36 cos^{2} (θ)

数字相乘:

36 \cdot 1 = 36

= - 36 + 36 cos^{2} (θ)

= 16 + 8 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ)

化简

16 + 8 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ) : 37 cos^{2} (θ) + 8 cos (θ) - 20

16 + 8 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ)

对同类项分组

= 8 cos (θ) + cos^{2} (θ) + 36 cos^{2} (θ) + 16 - 36

同类项相加:

cos^{2} (θ) + 36 cos^{2} (θ) = 37 cos^{2} (θ)

= 8 cos (θ) + 37 cos^{2} (θ) + 16 - 36

数字相加/相减:

16 - 36 = - 20

= 37 cos^{2} (θ) + 8 cos (θ) - 20

= 37 cos^{2} (θ) + 8 cos (θ) - 20

= 37 cos^{2} (θ) + 8 cos (θ) - 20

- 20 + 37 cos^{2} (θ) + 8 cos (θ) = 0

用替代法求解

- 20 + 37 cos^{2} (θ) + 8 cos (θ) = 0

令

: cos (θ) = u

- 20 + 37 u^{2} + 8 u = 0

- 20 + 37 u^{2} + 8 u = 0 : u = \frac{2 ( 3 21 - 2 )}{37}, u = - \frac{2 ( 2 + 3 21 )}{37}

- 20 + 37 u^{2} + 8 u = 0

改写成标准形式

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

37 u^{2} + 8 u - 20 = 0

使用求根公式求解

37 u^{2} + 8 u - 20 = 0

二次方程求根公式:

若

a = 37, b = 8, c = - 20

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 37 ( - 20 )}{2 \cdot 37}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 37 ( - 20 )}{2 \cdot 37}

8^{2} - 4 \cdot 37 (- 20) = 1221

8^{2} - 4 \cdot 37 (- 20)

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 37 \cdot 20

数字相乘:

4 \cdot 37 \cdot 20 = 2960

= 8^{2} + 2960

8^{2} = 64

= 64 + 2960

数字相加:

64 + 2960 = 3024

= 3024

3024

质因数分解:

2^{4} \cdot 3^{3} \cdot 7

3024

3024

除以

23024 = 1512 \cdot 2

= 2 \cdot 1512

1512

除以

21512 = 756 \cdot 2

= 2 \cdot 2 \cdot 756

756

除以

2756 = 378 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 378

378

除以

2378 = 189 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 189

189

除以

3189 = 63 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 63

63

除以

363 = 21 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 21

21

除以

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7

2, 3, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7

= 2^{4} \cdot 3^{3} \cdot 7

= 2^{4} \cdot 3^{3} \cdot 7

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 3^{2} \cdot 3 \cdot 7

使用根式运算法则:

n ab = n a n b

= 2^{4} 3^{2} 3 \cdot 7

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 3^{2} 3 \cdot 7

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{2} \cdot 3 3 \cdot 7

整理后得

= 1221

u_{1, 2} = \frac{- 8 \pm 12 21}{2 \cdot 37}

将解分隔开

u_{1} = \frac{- 8 + 12 21}{2 \cdot 37}, u_{2} = \frac{- 8 - 12 21}{2 \cdot 37}

u = \frac{- 8 + 12 21}{2 \cdot 37} : \frac{2 ( 3 21 - 2 )}{37}

\frac{- 8 + 12 21}{2 \cdot 37}

数字相乘:

2 \cdot 37 = 74

= \frac{- 8 + 12 21}{74}

分解

- 8 + 1221 : 4 (- 2 + 321)

- 8 + 1221

改写为

= - 4 \cdot 2 + 4 \cdot 321

因式分解出通项

4

= 4 (- 2 + 321)

= \frac{4 ( - 2 + 3 21 )}{74}

约分:

2

= \frac{2 ( 3 21 - 2 )}{37}

u = \frac{- 8 - 12 21}{2 \cdot 37} : - \frac{2 ( 2 + 3 21 )}{37}

\frac{- 8 - 12 21}{2 \cdot 37}

数字相乘:

2 \cdot 37 = 74

= \frac{- 8 - 12 21}{74}

分解

- 8 - 1221 : - 4 (2 + 321)

- 8 - 1221

改写为

= - 4 \cdot 2 - 4 \cdot 321

因式分解出通项

4

= - 4 (2 + 321)

= - \frac{4 ( 2 + 3 21 )}{74}

约分:

2

= - \frac{2 ( 2 + 3 21 )}{37}

二次方程组的解是:

u = \frac{2 ( 3 21 - 2 )}{37}, u = - \frac{2 ( 2 + 3 21 )}{37}

u = cos (θ)

代回

cos (θ) = \frac{2 ( 3 21 - 2 )}{37}, cos (θ) = - \frac{2 ( 2 + 3 21 )}{37}

cos (θ) = \frac{2 ( 3 21 - 2 )}{37}, cos (θ) = - \frac{2 ( 2 + 3 21 )}{37}

cos (θ) = \frac{2 ( 3 21 - 2 )}{37} : θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

cos (θ) = \frac{2 ( 3 21 - 2 )}{37}

使用反三角函数性质

cos (θ) = \frac{2 ( 3 21 - 2 )}{37}

cos (θ) = \frac{2 ( 3 21 - 2 )}{37}

的通解

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

θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

cos (θ) = - \frac{2 ( 2 + 3 21 )}{37} : θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

cos (θ) = - \frac{2 ( 2 + 3 21 )}{37}

使用反三角函数性质

cos (θ) = - \frac{2 ( 2 + 3 21 )}{37}

cos (θ) = - \frac{2 ( 2 + 3 21 )}{37}

的通解

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

θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

合并所有解

θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

将解代入原方程进行验证

将它们代入

cot (θ) + 4 csc (θ) = 6

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

的解:

真

arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1

对于

cot (θ) + 4 csc (θ) = 6

代入

θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1

cot (arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1) + 4 csc (arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1

对于

cot (θ) + 4 csc (θ) = 6

代入

θ = 2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1

cot (2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1) + 4 csc (2 π - arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 π 1) = 6

整理后得

- 6 = 6

\Rightarrow 假

检验

arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

的解:

真

arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1

对于

cot (θ) + 4 csc (θ) = 6

代入

θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1

cot (arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1) + 4 csc (arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

- arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

的解:

假

- arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1

对于

cot (θ) + 4 csc (θ) = 6

代入

θ = - arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1

cot (- arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1) + 4 csc (- arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 π 1) = 6

整理后得

- 6 = 6

\Rightarrow 假

θ = arccos (\frac{2 ( 3 21 - 2 )}{37}) + 2 πn, θ = arccos (- \frac{2 ( 2 + 3 21 )}{37}) + 2 πn

以小数形式表示解

θ = 0.88277 \dots + 2 πn, θ = 2.58911 \dots + 2 πn

作图

流行的例子

sin(x)sqrt(2)=-sin(x)

sin (x) 2 = - sin (x)

sin^2(x)-9cos(x)-9=0

sin^{2} (x) - 9 cos (x) - 9 = 0

sin (a) = \frac{8}{17}

cot (t) = 1

4 sin (x) + 3 = 5