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受欢迎的三角函数 >

3cot^2(x)-14csc(x)-2=0

解答

3 cot^{2} (x) - 14 csc (x) - 2 = 0

解答

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn

+1

度数

x = 11.53695 \dots^{\circ} + 36 0^{\circ} n, x = 168.46304 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cot^{2} (x) - 14 csc (x) - 2 = 0

使用三角恒等式改写

- 2 - 14 csc (x) + 3 cot^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 2 - 14 csc (x) + 3 (csc^{2} (x) - 1)

化简

- 2 - 14 csc (x) + 3 (csc^{2} (x) - 1) : 3 csc^{2} (x) - 14 csc (x) - 5

- 2 - 14 csc (x) + 3 (csc^{2} (x) - 1)

乘开

3 (csc^{2} (x) - 1) : 3 csc^{2} (x) - 3

3 (csc^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 3, b = csc^{2} (x), c = 1

= 3 csc^{2} (x) - 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 3 csc^{2} (x) - 3

= - 2 - 14 csc (x) + 3 csc^{2} (x) - 3

化简

- 2 - 14 csc (x) + 3 csc^{2} (x) - 3 : 3 csc^{2} (x) - 14 csc (x) - 5

- 2 - 14 csc (x) + 3 csc^{2} (x) - 3

对同类项分组

= - 14 csc (x) + 3 csc^{2} (x) - 2 - 3

数字相减:

- 2 - 3 = - 5

= 3 csc^{2} (x) - 14 csc (x) - 5

= 3 csc^{2} (x) - 14 csc (x) - 5

= 3 csc^{2} (x) - 14 csc (x) - 5

- 5 - 14 csc (x) + 3 csc^{2} (x) = 0

用替代法求解

- 5 - 14 csc (x) + 3 csc^{2} (x) = 0

令

: csc (x) = u

- 5 - 14 u + 3 u^{2} = 0

- 5 - 14 u + 3 u^{2} = 0 : u = 5, u = - \frac{1}{3}

- 5 - 14 u + 3 u^{2} = 0

改写成标准形式

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

3 u^{2} - 14 u - 5 = 0

使用求根公式求解

3 u^{2} - 14 u - 5 = 0

二次方程求根公式:

若

a = 3, b = - 14, c = - 5

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

(- 14)^{2} - 4 \cdot 3 (- 5) = 16

(- 14)^{2} - 4 \cdot 3 (- 5)

使用法则

- (- a) = a

= (- 14)^{2} + 4 \cdot 3 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 14)^{2} = 1 4^{2}

= 1 4^{2} + 4 \cdot 3 \cdot 5

数字相乘:

4 \cdot 3 \cdot 5 = 60

= 1 4^{2} + 60

1 4^{2} = 196

= 196 + 60

数字相加:

196 + 60 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 14 ) \pm 16}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 14 ) + 16}{2 \cdot 3}, u_{2} = \frac{- ( - 14 ) - 16}{2 \cdot 3}

u = \frac{- ( - 14 ) + 16}{2 \cdot 3} : 5

\frac{- ( - 14 ) + 16}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{14 + 16}{2 \cdot 3}

数字相加:

14 + 16 = 30

= \frac{30}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{30}{6}

数字相除:

\frac{30}{6} = 5

= 5

u = \frac{- ( - 14 ) - 16}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 14 ) - 16}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{14 - 16}{2 \cdot 3}

数字相减:

14 - 16 = - 2

= \frac{- 2}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 2}{6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

二次方程组的解是:

u = 5, u = - \frac{1}{3}

u = csc (x)

代回

csc (x) = 5, csc (x) = - \frac{1}{3}

csc (x) = 5, csc (x) = - \frac{1}{3}

csc (x) = 5 : x = arccsc (5) + 2 πn, x = π - arccsc (5) + 2 πn

csc (x) = 5

使用反三角函数性质

csc (x) = 5

csc (x) = 5

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (5) + 2 πn, x = π - arccsc (5) + 2 πn

x = arccsc (5) + 2 πn, x = π - arccsc (5) + 2 πn

csc (x) = - \frac{1}{3} :

无解

csc (x) = - \frac{1}{3}

csc (x) \leq - 1 or csc (x) \geq 1

无解

合并所有解

x = arccsc (5) + 2 πn, x = π - arccsc (5) + 2 πn

以小数形式表示解

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn

作图

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