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tan(x)cot(x)-tan(x)+2cot(x)=0

解答

tan (x) cot (x) - tan (x) + 2 cot (x) = 0

解答

x = 1.10714 \dots + πn, x = \frac{3 π}{4} + πn

+1

度数

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) cot (x) - tan (x) + 2 cot (x) = 0

使用三角恒等式改写

- tan (x) + 2 cot (x) + cot (x) tan (x)

使用基本三角恒等式:

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

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

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

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

分式相乘:

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

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

约分:

cot (x)

= 1

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

1 - \frac{1}{cot ( x )} + 2 cot (x) = 0

用替代法求解

1 - \frac{1}{cot ( x )} + 2 cot (x) = 0

令

: cot (x) = u

1 - \frac{1}{u} + 2 u = 0

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

1 - \frac{1}{u} + 2 u = 0

在两边乘以

u

1 - \frac{1}{u} + 2 u = 0

在两边乘以

u

1 \cdot u - \frac{1}{u} u + 2 uu = 0 \cdot u

化简

1 \cdot u - \frac{1}{u} u + 2 uu = 0 \cdot u

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= 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

2 uu : 2 u^{2}

2 uu

使用指数法则:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 1, c = - 1

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

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

1^{2} - 4 \cdot 2 (- 1) = 3

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

1 - \frac{1}{u} + 2 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 = - 1

u = cot (x)

代回

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

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

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) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

cot (x) = - 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = arccot (\frac{1}{2}) + πn, x = \frac{3 π}{4} + πn

以小数形式表示解

x = 1.10714 \dots + πn, x = \frac{3 π}{4} + πn

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