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受欢迎的三角函数 >

tan(x)+cot(x)=5

解答

tan (x) + cot (x) = 5

解答

x = 0.20575 \dots + πn, x = 1.36503 \dots + πn

+1

度数

x = 11.78908 \dots^{\circ} + 18 0^{\circ} n, x = 78.21091 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) + cot (x) = 5

两边减去

5

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: cot (x) = u

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

- 5 + u + \frac{1}{u} = 0 : u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

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

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 5 u + u^{2} + 1 = 0 : u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 5)^{2} - 4 \cdot 1 \cdot 1 = 21

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 5^{2} - 4

5^{2} = 25

= 25 - 4

数字相减:

25 - 4 = 21

= 21

u_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

u = \frac{- ( - 5 ) + 21}{2 \cdot 1} : \frac{5 + 21}{2}

\frac{- ( - 5 ) + 21}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 + 21}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 + 21}{2}

u = \frac{- ( - 5 ) - 21}{2 \cdot 1} : \frac{5 - 21}{2}

\frac{- ( - 5 ) - 21}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 - 21}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 - 21}{2}

二次方程组的解是:

u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

u = \frac{5 + 21}{2}, u = \frac{5 - 21}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

- 5 + u + \frac{1}{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{5 + 21}{2}, u = \frac{5 - 21}{2}

u = cot (x)

代回

cot (x) = \frac{5 + 21}{2}, cot (x) = \frac{5 - 21}{2}

cot (x) = \frac{5 + 21}{2}, cot (x) = \frac{5 - 21}{2}

cot (x) = \frac{5 + 21}{2} : x = arccot (\frac{5 + 21}{2}) + πn

cot (x) = \frac{5 + 21}{2}

使用反三角函数性质

cot (x) = \frac{5 + 21}{2}

cot (x) = \frac{5 + 21}{2}

的通解

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

x = arccot (\frac{5 + 21}{2}) + πn

x = arccot (\frac{5 + 21}{2}) + πn

cot (x) = \frac{5 - 21}{2} : x = arccot (\frac{5 - 21}{2}) + πn

cot (x) = \frac{5 - 21}{2}

使用反三角函数性质

cot (x) = \frac{5 - 21}{2}

cot (x) = \frac{5 - 21}{2}

的通解

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

x = arccot (\frac{5 - 21}{2}) + πn

x = arccot (\frac{5 - 21}{2}) + πn

合并所有解

x = arccot (\frac{5 + 21}{2}) + πn, x = arccot (\frac{5 - 21}{2}) + πn

以小数形式表示解

x = 0.20575 \dots + πn, x = 1.36503 \dots + πn

作图

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