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受欢迎的三角函数 >

cot(θ)+2csc(θ)=6

解答

cot (θ) + 2 csc (θ) = 6

解答

θ = 0.50017 \dots + 2 πn, θ = 2.97171 \dots + 2 πn

+1

度数

θ = 28.65815 \dots^{\circ} + 36 0^{\circ} n, θ = 170.26648 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cot (θ) + 2 csc (θ) = 6

两边减去

6

cot (θ) + 2 csc (θ) - 6 = 0

用 sin, cos 表示

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 6 = 0

化简

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 6 : \frac{cos ( θ ) + 2 - 6 sin ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 6

2 \cdot \frac{1}{sin ( θ )} = \frac{2}{sin ( θ )}

2 \cdot \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot 2}{sin ( θ )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} - 6

合并分式

\frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} : \frac{cos ( θ ) + 2}{sin ( θ )}

使用法则

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

= \frac{cos ( θ ) + 2}{sin ( θ )}

= \frac{cos ( θ ) + 2}{sin ( θ )} - 6

将项转换为分式:

6 = \frac{6 sin ( θ )}{sin ( θ )}

= \frac{cos ( θ ) + 2}{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 ) + 2 - 6 sin ( θ )}{sin ( θ )}

\frac{cos ( θ ) + 2 - 6 sin ( θ )}{sin ( θ )} = 0

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

cos (θ) + 2 - 6 sin (θ) = 0

两边加上

6 sin (θ)

cos (θ) + 2 = 6 sin (θ)

两边进行平方

(cos (θ) + 2)^{2} = (6 sin (θ))^{2}

两边减去

(6 sin (θ))^{2}

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

使用三角恒等式改写

(2 + 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)

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

化简

(2 + cos (θ))^{2} - 36 (1 - cos^{2} (θ)) : 37 cos^{2} (θ) + 4 cos (θ) - 32

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

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

使用完全平方公式:

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

a = 2, b = cos (θ)

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

化简

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

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

2^{2} = 4

= 4 + 2 \cdot 2 cos (θ) + cos^{2} (θ)

数字相乘:

2 \cdot 2 = 4

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 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)

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

化简

4 + 4 cos (θ) + cos^{2} (θ) - 36 + 36 cos^{2} (θ) : 37 cos^{2} (θ) + 4 cos (θ) - 32

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

对同类项分组

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

同类项相加:

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

= 4 cos (θ) + 37 cos^{2} (θ) + 4 - 36

数字相加/相减:

4 - 36 = - 32

= 37 cos^{2} (θ) + 4 cos (θ) - 32

= 37 cos^{2} (θ) + 4 cos (θ) - 32

= 37 cos^{2} (θ) + 4 cos (θ) - 32

- 32 + 37 cos^{2} (θ) + 4 cos (θ) = 0

用替代法求解

- 32 + 37 cos^{2} (θ) + 4 cos (θ) = 0

令

: cos (θ) = u

- 32 + 37 u^{2} + 4 u = 0

- 32 + 37 u^{2} + 4 u = 0 : u = \frac{2 ( 3 33 - 1 )}{37}, u = - \frac{2 ( 1 + 3 33 )}{37}

- 32 + 37 u^{2} + 4 u = 0

改写成标准形式

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

37 u^{2} + 4 u - 32 = 0

使用求根公式求解

37 u^{2} + 4 u - 32 = 0

二次方程求根公式:

若

a = 37, b = 4, c = - 32

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 37 ( - 32 )}{2 \cdot 37}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 37 ( - 32 )}{2 \cdot 37}

4^{2} - 4 \cdot 37 (- 32) = 1233

4^{2} - 4 \cdot 37 (- 32)

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 37 \cdot 32

数字相乘:

4 \cdot 37 \cdot 32 = 4736

= 4^{2} + 4736

4^{2} = 16

= 16 + 4736

数字相加:

16 + 4736 = 4752

= 4752

4752

质因数分解:

2^{4} \cdot 3^{3} \cdot 11

4752

4752

除以

24752 = 2376 \cdot 2

= 2 \cdot 2376

2376

除以

22376 = 1188 \cdot 2

= 2 \cdot 2 \cdot 1188

1188

除以

21188 = 594 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 594

594

除以

2594 = 297 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 297

297

除以

3297 = 99 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 99

99

除以

399 = 33 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 33

33

除以

333 = 11 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 11

2, 3, 11

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 11

= 2^{4} \cdot 3^{3} \cdot 11

= 2^{4} \cdot 3^{3} \cdot 11

使用指数法则:

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

= 2^{4} \cdot 3^{2} \cdot 3 \cdot 11

使用根式运算法则:

n ab = n a n b

= 2^{4} 3^{2} 3 \cdot 11

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 11

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2^{2} \cdot 3 3 \cdot 11

整理后得

= 1233

u_{1, 2} = \frac{- 4 \pm 12 33}{2 \cdot 37}

将解分隔开

u_{1} = \frac{- 4 + 12 33}{2 \cdot 37}, u_{2} = \frac{- 4 - 12 33}{2 \cdot 37}

u = \frac{- 4 + 12 33}{2 \cdot 37} : \frac{2 ( 3 33 - 1 )}{37}

\frac{- 4 + 12 33}{2 \cdot 37}

数字相乘:

2 \cdot 37 = 74

= \frac{- 4 + 12 33}{74}

分解

- 4 + 1233 : 4 (- 1 + 333)

- 4 + 1233

改写为

= - 4 \cdot 1 + 4 \cdot 333

因式分解出通项

4

= 4 (- 1 + 333)

= \frac{4 ( - 1 + 3 33 )}{74}

约分:

2

= \frac{2 ( 3 33 - 1 )}{37}

u = \frac{- 4 - 12 33}{2 \cdot 37} : - \frac{2 ( 1 + 3 33 )}{37}

\frac{- 4 - 12 33}{2 \cdot 37}

数字相乘:

2 \cdot 37 = 74

= \frac{- 4 - 12 33}{74}

分解

- 4 - 1233 : - 4 (1 + 333)

- 4 - 1233

改写为

= - 4 \cdot 1 - 4 \cdot 333

因式分解出通项

4

= - 4 (1 + 333)

= - \frac{4 ( 1 + 3 33 )}{74}

约分:

2

= - \frac{2 ( 1 + 3 33 )}{37}

二次方程组的解是:

u = \frac{2 ( 3 33 - 1 )}{37}, u = - \frac{2 ( 1 + 3 33 )}{37}

u = cos (θ)

代回

cos (θ) = \frac{2 ( 3 33 - 1 )}{37}, cos (θ) = - \frac{2 ( 1 + 3 33 )}{37}

cos (θ) = \frac{2 ( 3 33 - 1 )}{37}, cos (θ) = - \frac{2 ( 1 + 3 33 )}{37}

cos (θ) = \frac{2 ( 3 33 - 1 )}{37} : θ = arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

cos (θ) = \frac{2 ( 3 33 - 1 )}{37}

使用反三角函数性质

cos (θ) = \frac{2 ( 3 33 - 1 )}{37}

cos (θ) = \frac{2 ( 3 33 - 1 )}{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 33 - 1 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

θ = arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

cos (θ) = - \frac{2 ( 1 + 3 33 )}{37} : θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

cos (θ) = - \frac{2 ( 1 + 3 33 )}{37}

使用反三角函数性质

cos (θ) = - \frac{2 ( 1 + 3 33 )}{37}

cos (θ) = - \frac{2 ( 1 + 3 33 )}{37}

的通解

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

θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

合并所有解

θ = arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn, θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn, θ = - arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

将解代入原方程进行验证

将它们代入

cot (θ) + 2 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 33 - 1 )}{37}) + 2 πn

的解:

真

arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

代入

n = 1

arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1

对于

cot (θ) + 2 csc (θ) = 6

代入

θ = arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1

cot (arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1) + 2 csc (arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

的解:

假

2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1

对于

cot (θ) + 2 csc (θ) = 6

代入

θ = 2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1

cot (2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1) + 2 csc (2 π - arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 π 1) = 6

整理后得

- 6 = 6

\Rightarrow 假

检验

arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

的解:

真

arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

代入

n = 1

arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1

对于

cot (θ) + 2 csc (θ) = 6

代入

θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1

cot (arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1) + 2 csc (arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1) = 6

整理后得

6 = 6

\Rightarrow 真

检验

- arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

的解:

假

- arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

代入

n = 1

- arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1

对于

cot (θ) + 2 csc (θ) = 6

代入

θ = - arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1

cot (- arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1) + 2 csc (- arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 π 1) = 6

整理后得

- 6 = 6

\Rightarrow 假

θ = arccos (\frac{2 ( 3 33 - 1 )}{37}) + 2 πn, θ = arccos (- \frac{2 ( 1 + 3 33 )}{37}) + 2 πn

以小数形式表示解

θ = 0.50017 \dots + 2 πn, θ = 2.97171 \dots + 2 πn

作图

流行的例子

(1+cot(x))/(1+tan(x))=5

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

4 = 4 cos (x)

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

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

3sec^2(u)+7tan(u)=3

3 sec^{2} (u) + 7 tan (u) = 3

3sin(t)=2cos^2(t)

3 sin (t) = 2 cos^{2} (t)