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受欢迎的三角函数 >

(1+tan(x))/(1+cot(x))=2

解答

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

解答

x = 1.10714 \dots + πn

+1

度数

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

求解步骤

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

两边减去

2

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

化简

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

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

将项转换为分式:

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

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

乘开

1 + tan (x) - 2 (1 + cot (x)) : tan (x) - 2 cot (x) - 1

1 + tan (x) - 2 (1 + cot (x))

乘开

- 2 (1 + cot (x)) : - 2 - 2 cot (x)

- 2 (1 + cot (x))

使用分配律:

a (b + c) = ab + a c

a = - 2, b = 1, c = cot (x)

= - 2 \cdot 1 + (- 2) cot (x)

使用加减运算法则

+ (- a) = - a

= - 2 \cdot 1 - 2 cot (x)

数字相乘:

2 \cdot 1 = 2

= - 2 - 2 cot (x)

= 1 + tan (x) - 2 - 2 cot (x)

化简

1 + tan (x) - 2 - 2 cot (x) : tan (x) - 2 cot (x) - 1

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

对同类项分组

= tan (x) - 2 cot (x) + 1 - 2

数字相加/相减:

1 - 2 = - 1

= tan (x) - 2 cot (x) - 1

= tan (x) - 2 cot (x) - 1

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

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

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

tan (x) - 2 cot (x) - 1 = 0

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 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:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

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

数字相加:

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

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

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

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

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

u = cot (x)

代回

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

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

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

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

合并所有解

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

因为方程对以下值无定义:

\frac{3 π}{4} + πn

x = arccot (\frac{1}{2}) + πn

以小数形式表示解

x = 1.10714 \dots + πn

作图

流行的例子

tan(x)+1= 1/(sqrt(3))+1/(sqrt(3))cot(x)

tan (x) + 1 = \frac{1}{3} + \frac{1}{3} cot (x)

sin (8 x) = 1

2cos^2(w)-3cos(w)-5=0

2 cos^{2} (w) - 3 cos (w) - 5 = 0

4 sin (x) = csc (x)

cot (θ) = - \frac{3}{2}