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受欢迎的三角函数 >

tan(x)+cot(x)= 5/2

解答

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

解答

x = 0.46364 \dots + πn, x = 1.10714 \dots + πn

+1

度数

x = 26.56505 \dots^{\circ} + 18 0^{\circ} n, x = 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

\frac{5}{2}

tan (x) + cot (x) - \frac{5}{2} = 0

化简

tan (x) + cot (x) - \frac{5}{2} : \frac{2 tan ( x ) + 2 cot ( x ) - 5}{2}

tan (x) + cot (x) - \frac{5}{2}

将项转换为分式:

tan (x) = \frac{tan ( x ) 2}{2}, cot (x) = \frac{cot ( x ) 2}{2}

= \frac{tan ( x ) \cdot 2}{2} + \frac{cot ( x ) \cdot 2}{2} - \frac{5}{2}

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{tan ( x ) \cdot 2 + cot ( x ) \cdot 2 - 5}{2}

\frac{2 tan ( x ) + 2 cot ( x ) - 5}{2} = 0

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

2 tan (x) + 2 cot (x) - 5 = 0

使用三角恒等式改写

- 5 + 2 cot (x) + 2 tan (x)

使用基本三角恒等式:

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

= - 5 + 2 cot (x) + 2 \cdot \frac{1}{cot ( x )}

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

\frac{2}{u} u : 2

\frac{2}{u} u

分式相乘:

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

= \frac{2 u}{u}

约分:

u

= 2

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

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = - 5, c = 2

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

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

(- 5)^{2} - 4 \cdot 2 \cdot 2 = 3

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 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{- ( - 5 ) \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

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

\frac{- ( - 5 ) + 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 + 3}{2 \cdot 2}

数字相加:

5 + 3 = 8

= \frac{8}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

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

\frac{- ( - 5 ) - 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 - 3}{2 \cdot 2}

数字相减:

5 - 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = cot (x)

代回

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

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

cot (x) = 2 : x = arccot (2) + πn

cot (x) = 2

使用反三角函数性质

cot (x) = 2

cot (x) = 2

的通解

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

x = arccot (2) + πn

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

以小数形式表示解

x = 0.46364 \dots + πn, x = 1.10714 \dots + πn

作图

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