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受欢迎的三角函数 >

6cot(x)=5-tan(x)

解答

6 cot (x) = 5 - tan (x)

解答

x = 1.10714 \dots + πn, x = 1.24904 \dots + πn

+1

度数

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 71.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

6 cot (x) = 5 - tan (x)

两边减去

5 - tan (x)

6 cot (x) - 5 + tan (x) = 0

使用三角恒等式改写

- 5 + tan (x) + 6 cot (x)

使用基本三角恒等式:

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

= - 5 + \frac{1}{cot ( x )} + 6 cot (x)

- 5 + \frac{1}{cot ( x )} + 6 cot (x) = 0

用替代法求解

- 5 + \frac{1}{cot ( x )} + 6 cot (x) = 0

令

: cot (x) = u

- 5 + \frac{1}{u} + 6 u = 0

- 5 + \frac{1}{u} + 6 u = 0 : u = \frac{1}{2}, u = \frac{1}{3}

- 5 + \frac{1}{u} + 6 u = 0

在两边乘以

u

- 5 + \frac{1}{u} + 6 u = 0

在两边乘以

u

- 5 u + \frac{1}{u} u + 6 uu = 0 \cdot u

化简

- 5 u + \frac{1}{u} u + 6 uu = 0 \cdot u

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

6 uu : 6 u^{2}

6 uu

使用指数法则:

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

uu = u^{1 + 1}

= 6 u^{1 + 1}

数字相加:

1 + 1 = 2

= 6 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 5 u + 1 + 6 u^{2} = 0

- 5 u + 1 + 6 u^{2} = 0

- 5 u + 1 + 6 u^{2} = 0

解

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

- 5 u + 1 + 6 u^{2} = 0

改写成标准形式

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

6 u^{2} - 5 u + 1 = 0

使用求根公式求解

6 u^{2} - 5 u + 1 = 0

二次方程求根公式:

若

a = 6, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 1 = 1

(- 5)^{2} - 4 \cdot 6 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 5^{2} - 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

u = \frac{- ( - 5 ) + 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

数字相加:

5 + 1 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

数字相减:

5 - 1 = 4

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

数字相乘:

2 \cdot 6 = 12

= \frac{4}{12}

约分:

4

= \frac{1}{3}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

- 5 + \frac{1}{u} + 6 u

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = cot (x)

代回

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

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

cot (x) = \frac{1}{2} : x = arccot (\frac{1}{2}) + πn

cot (x) = \frac{1}{2}

使用反三角函数性质

cot (x) = \frac{1}{2}

cot (x) = \frac{1}{2}

的通解

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

x = arccot (\frac{1}{2}) + πn

x = arccot (\frac{1}{2}) + πn

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

合并所有解

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

以小数形式表示解

x = 1.10714 \dots + πn, x = 1.24904 \dots + πn

作图

流行的例子

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

((sin^2(x)))/((1-cos(x)))=1.23

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