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受欢迎的三角函数 >

-6csc^2(x)+10cot(x)=-10

解答

- 6 csc^{2} (x) + 10 cot (x) = - 10

解答

x = 1.89254 \dots + πn, x = 0.46364 \dots + πn

+1

度数

x = 108.43494 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

- 6 csc^{2} (x) + 10 cot (x) = - 10

两边减去

- 10

- 6 csc^{2} (x) + 10 cot (x) + 10 = 0

使用三角恒等式改写

10 + 10 cot (x) - 6 csc^{2} (x)

使用毕达哥拉斯恒等式:

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

= 10 + 10 cot (x) - 6 (1 + cot^{2} (x))

化简

10 + 10 cot (x) - 6 (1 + cot^{2} (x)) : 10 cot (x) - 6 cot^{2} (x) + 4

10 + 10 cot (x) - 6 (1 + cot^{2} (x))

乘开

- 6 (1 + cot^{2} (x)) : - 6 - 6 cot^{2} (x)

- 6 (1 + cot^{2} (x))

使用分配律:

a (b + c) = ab + a c

a = - 6, b = 1, c = cot^{2} (x)

= - 6 \cdot 1 + (- 6) cot^{2} (x)

使用加减运算法则

+ (- a) = - a

= - 6 \cdot 1 - 6 cot^{2} (x)

数字相乘:

6 \cdot 1 = 6

= - 6 - 6 cot^{2} (x)

= 10 + 10 cot (x) - 6 - 6 cot^{2} (x)

化简

10 + 10 cot (x) - 6 - 6 cot^{2} (x) : 10 cot (x) - 6 cot^{2} (x) + 4

10 + 10 cot (x) - 6 - 6 cot^{2} (x)

对同类项分组

= 10 cot (x) - 6 cot^{2} (x) + 10 - 6

数字相加/相减:

10 - 6 = 4

= 10 cot (x) - 6 cot^{2} (x) + 4

= 10 cot (x) - 6 cot^{2} (x) + 4

= 10 cot (x) - 6 cot^{2} (x) + 4

4 + 10 cot (x) - 6 cot^{2} (x) = 0

用替代法求解

4 + 10 cot (x) - 6 cot^{2} (x) = 0

令

: cot (x) = u

4 + 10 u - 6 u^{2} = 0

4 + 10 u - 6 u^{2} = 0 : u = - \frac{1}{3}, u = 2

4 + 10 u - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} + 10 u + 4 = 0

使用求根公式求解

- 6 u^{2} + 10 u + 4 = 0

二次方程求根公式:

若

a = - 6, b = 10, c = 4

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 6 ) \cdot 4}{2 ( - 6 )}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 6 ) \cdot 4}{2 ( - 6 )}

1 0^{2} - 4 (- 6) \cdot 4 = 14

1 0^{2} - 4 (- 6) \cdot 4

使用法则

- (- a) = a

= 1 0^{2} + 4 \cdot 6 \cdot 4

数字相乘:

4 \cdot 6 \cdot 4 = 96

= 1 0^{2} + 96

1 0^{2} = 100

= 100 + 96

数字相加:

100 + 96 = 196

= 196

因式分解数字:

196 = 1 4^{2}

= 1 4^{2}

使用根式运算法则:

n a^{n} = a

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- 10 \pm 14}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 10 + 14}{2 ( - 6 )}, u_{2} = \frac{- 10 - 14}{2 ( - 6 )}

u = \frac{- 10 + 14}{2 ( - 6 )} : - \frac{1}{3}

\frac{- 10 + 14}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 10 + 14}{- 2 \cdot 6}

数字相加/相减:

- 10 + 14 = 4

= \frac{4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4}{- 12}

使用分式法则:

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

= - \frac{4}{12}

约分:

4

= - \frac{1}{3}

u = \frac{- 10 - 14}{2 ( - 6 )} : 2

\frac{- 10 - 14}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 10 - 14}{- 2 \cdot 6}

数字相减:

- 10 - 14 = - 24

= \frac{- 24}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 24}{- 12}

使用分式法则:

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

= \frac{24}{12}

数字相除:

\frac{24}{12} = 2

= 2

二次方程组的解是:

u = - \frac{1}{3}, u = 2

u = cot (x)

代回

cot (x) = - \frac{1}{3}, cot (x) = 2

cot (x) = - \frac{1}{3}, cot (x) = 2

cot (x) = - \frac{1}{3} : x = arccot (- \frac{1}{3}) + πn

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

使用反三角函数性质

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

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

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- \frac{1}{3}) + πn

x = arccot (- \frac{1}{3}) + πn

cot (x) = 2 : x = arccot (2) + πn

cot (x) = 2

使用反三角函数性质

cot (x) = 2

cot (x) = 2

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (2) + πn

x = arccot (2) + πn

合并所有解

x = arccot (- \frac{1}{3}) + πn, x = arccot (2) + πn

以小数形式表示解

x = 1.89254 \dots + πn, x = 0.46364 \dots + πn

作图

流行的例子

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