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受欢迎的三角函数 >

tan(x)=4-3cot(x)

解答

tan (x) = 4 - 3 cot (x)

解答

x = \frac{π}{4} + πn, x = 1.24904 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 71.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) = 4 - 3 cot (x)

两边减去

4 - 3 cot (x)

tan (x) - 4 + 3 cot (x) = 0

使用三角恒等式改写

- 4 + tan (x) + 3 cot (x)

使用基本三角恒等式:

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

= - 4 + \frac{1}{cot ( x )} + 3 cot (x)

- 4 + \frac{1}{cot ( x )} + 3 cot (x) = 0

用替代法求解

- 4 + \frac{1}{cot ( x )} + 3 cot (x) = 0

令

: cot (x) = u

- 4 + \frac{1}{u} + 3 u = 0

- 4 + \frac{1}{u} + 3 u = 0 : u = 1, u = \frac{1}{3}

- 4 + \frac{1}{u} + 3 u = 0

在两边乘以

u

- 4 + \frac{1}{u} + 3 u = 0

在两边乘以

u

- 4 u + \frac{1}{u} u + 3 uu = 0 \cdot u

化简

- 4 u + \frac{1}{u} u + 3 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

3 uu : 3 u^{2}

3 uu

使用指数法则:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

数字相加:

1 + 1 = 2

= 3 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 4 u + 1 + 3 u^{2} = 0

- 4 u + 1 + 3 u^{2} = 0

- 4 u + 1 + 3 u^{2} = 0

解

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

- 4 u + 1 + 3 u^{2} = 0

改写成标准形式

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

3 u^{2} - 4 u + 1 = 0

使用求根公式求解

3 u^{2} - 4 u + 1 = 0

二次方程求根公式:

若

a = 3, b = - 4, c = 1

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数字相减:

16 - 12 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

将解分隔开

u_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 3}, u_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 3}

u = \frac{- ( - 4 ) + 2}{2 \cdot 3} : 1

\frac{- ( - 4 ) + 2}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{4 + 2}{2 \cdot 3}

数字相加:

4 + 2 = 6

= \frac{6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{6}{6}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 4 ) - 2}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 4 ) - 2}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{4 - 2}{2 \cdot 3}

数字相减:

4 - 2 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

- 4 + \frac{1}{u} + 3 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}{3}

u = cot (x)

代回

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

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

cot (x) = 1 : x = \frac{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{π}{4} + πn

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

以小数形式表示解

x = \frac{π}{4} + πn, x = 1.24904 \dots + πn

作图

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