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受欢迎的三角函数 >

3+4cot(x)=5csc(x)

解答

3 + 4 cot (x) = 5 csc (x)

解答

x = 0.64350 \dots + 2 πn

+1

度数

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 + 4 cot (x) = 5 csc (x)

两边减去

5 csc (x)

3 + 4 cot (x) - 5 csc (x) = 0

用 sin, cos 表示

3 + 4 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{sin ( x )} = 0

化简

3 + 4 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{sin ( x )} : \frac{3 sin ( x ) + 4 cos ( x ) - 5}{sin ( x )}

3 + 4 \cdot \frac{cos ( x )}{sin ( x )} - 5 \cdot \frac{1}{sin ( x )}

4 \cdot \frac{cos ( x )}{sin ( x )} = \frac{4 cos ( x )}{sin ( x )}

4 \cdot \frac{cos ( x )}{sin ( x )}

分式相乘:

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

= \frac{cos ( x ) \cdot 4}{sin ( x )}

5 \cdot \frac{1}{sin ( x )} = \frac{5}{sin ( x )}

5 \cdot \frac{1}{sin ( x )}

分式相乘:

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

= \frac{1 \cdot 5}{sin ( x )}

数字相乘:

1 \cdot 5 = 5

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

= 3 + \frac{4 cos ( x )}{sin ( x )} - \frac{5}{sin ( x )}

合并分式

\frac{4 cos ( x )}{sin ( x )} - \frac{5}{sin ( x )} : \frac{4 cos ( x ) - 5}{sin ( x )}

使用法则

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

= \frac{4 cos ( x ) - 5}{sin ( x )}

= 3 + \frac{4 cos ( x ) - 5}{sin ( x )}

将项转换为分式:

3 = \frac{3 sin ( x )}{sin ( x )}

= \frac{3 sin ( x )}{sin ( x )} + \frac{cos ( x ) \cdot 4 - 5}{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{3 sin ( x ) + cos ( x ) \cdot 4 - 5}{sin ( x )}

\frac{3 sin ( x ) + 4 cos ( x ) - 5}{sin ( x )} = 0

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

3 sin (x) + 4 cos (x) - 5 = 0

两边减去

4 cos (x)

3 sin (x) - 5 = - 4 cos (x)

两边进行平方

(3 sin (x) - 5)^{2} = (- 4 cos (x))^{2}

两边减去

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

(3 sin (x) - 5)^{2} - 16 cos^{2} (x) = 0

使用三角恒等式改写

(- 5 + 3 sin (x))^{2} - 16 cos^{2} (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)

= (- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x))

化简

(- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x)) : 25 sin^{2} (x) - 30 sin (x) + 9

(- 5 + 3 sin (x))^{2} - 16 (1 - sin^{2} (x))

(- 5 + 3 sin (x))^{2} : 25 - 30 sin (x) + 9 sin^{2} (x)

使用完全平方公式:

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

a = - 5, b = 3 sin (x)

= (- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2}

化简

(- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2} : 25 - 30 sin (x) + 9 sin^{2} (x)

(- 5)^{2} + 2 (- 5) \cdot 3 sin (x) + (3 sin (x))^{2}

去除括号:

(- a) = - a

= (- 5)^{2} - 2 \cdot 5 \cdot 3 sin (x) + (3 sin (x))^{2}

(- 5)^{2} = 25

(- 5)^{2}

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2}

5^{2} = 25

= 25

2 \cdot 5 \cdot 3 sin (x) = 30 sin (x)

2 \cdot 5 \cdot 3 sin (x)

数字相乘:

2 \cdot 5 \cdot 3 = 30

= 30 sin (x)

(3 sin (x))^{2} = 9 sin^{2} (x)

(3 sin (x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} sin^{2} (x)

3^{2} = 9

= 9 sin^{2} (x)

= 25 - 30 sin (x) + 9 sin^{2} (x)

= 25 - 30 sin (x) + 9 sin^{2} (x)

= 25 - 30 sin (x) + 9 sin^{2} (x) - 16 (1 - sin^{2} (x))

乘开

- 16 (1 - sin^{2} (x)) : - 16 + 16 sin^{2} (x)

- 16 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (x)

= - 16 \cdot 1 - (- 16) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (x)

数字相乘:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (x)

= 25 - 30 sin (x) + 9 sin^{2} (x) - 16 + 16 sin^{2} (x)

化简

25 - 30 sin (x) + 9 sin^{2} (x) - 16 + 16 sin^{2} (x) : 25 sin^{2} (x) - 30 sin (x) + 9

25 - 30 sin (x) + 9 sin^{2} (x) - 16 + 16 sin^{2} (x)

对同类项分组

= - 30 sin (x) + 9 sin^{2} (x) + 16 sin^{2} (x) + 25 - 16

同类项相加:

9 sin^{2} (x) + 16 sin^{2} (x) = 25 sin^{2} (x)

= - 30 sin (x) + 25 sin^{2} (x) + 25 - 16

数字相加/相减:

25 - 16 = 9

= 25 sin^{2} (x) - 30 sin (x) + 9

= 25 sin^{2} (x) - 30 sin (x) + 9

= 25 sin^{2} (x) - 30 sin (x) + 9

9 + 25 sin^{2} (x) - 30 sin (x) = 0

用替代法求解

9 + 25 sin^{2} (x) - 30 sin (x) = 0

令

: sin (x) = u

9 + 25 u^{2} - 30 u = 0

9 + 25 u^{2} - 30 u = 0 : u = \frac{3}{5}

9 + 25 u^{2} - 30 u = 0

改写成标准形式

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

25 u^{2} - 30 u + 9 = 0

使用求根公式求解

25 u^{2} - 30 u + 9 = 0

二次方程求根公式:

若

a = 25, b = - 30, c = 9

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 25 \cdot 9}{2 \cdot 25}

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 25 \cdot 9}{2 \cdot 25}

(- 30)^{2} - 4 \cdot 25 \cdot 9 = 0

(- 30)^{2} - 4 \cdot 25 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 30)^{2} = 3 0^{2}

= 3 0^{2} - 4 \cdot 25 \cdot 9

数字相乘:

4 \cdot 25 \cdot 9 = 900

= 3 0^{2} - 900

3 0^{2} = 900

= 900 - 900

数字相减:

900 - 900 = 0

= 0

u_{1, 2} = \frac{- ( - 30 ) \pm 0}{2 \cdot 25}

u = \frac{- ( - 30 )}{2 \cdot 25}

\frac{- ( - 30 )}{2 \cdot 25} = \frac{3}{5}

\frac{- ( - 30 )}{2 \cdot 25}

使用法则

- (- a) = a

= \frac{30}{2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{30}{50}

约分:

10

= \frac{3}{5}

u = \frac{3}{5}

二次方程组的解是:

u = \frac{3}{5}

u = sin (x)

代回

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

使用反三角函数性质

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

的通解

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

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

合并所有解

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

将解代入原方程进行验证

将它们代入

3 + 4 cot (x) = 5 csc (x)

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{3}{5}) + 2 πn

的解:

真

arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

arcsin (\frac{3}{5}) + 2 π 1

对于

3 + 4 cot (x) = 5 csc (x)

代入

x = arcsin (\frac{3}{5}) + 2 π 1

3 + 4 cot (arcsin (\frac{3}{5}) + 2 π 1) = 5 csc (arcsin (\frac{3}{5}) + 2 π 1)

整理后得

8.33333 \dots = 8.33333 \dots

\Rightarrow 真

检验

π - arcsin (\frac{3}{5}) + 2 πn

的解:

假

π - arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

对于

3 + 4 cot (x) = 5 csc (x)

代入

x = π - arcsin (\frac{3}{5}) + 2 π 1

3 + 4 cot (π - arcsin (\frac{3}{5}) + 2 π 1) = 5 csc (π - arcsin (\frac{3}{5}) + 2 π 1)

整理后得

- 2.33333 \dots = 8.33333 \dots

\Rightarrow 假

x = arcsin (\frac{3}{5}) + 2 πn

以小数形式表示解

x = 0.64350 \dots + 2 πn

作图

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