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

解答

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

解答

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

+1

度数

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

求解步骤

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

化简

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

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

将项转换为分式:

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

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

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

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} tan (x)

分式相乘:

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

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

2 tan (x) tan (x) = 2 tan^{2} (x)

2 tan (x) tan (x)

使用指数法则:

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

tan (x) tan (x) = tan^{1 + 1} (x)

= 2 tan^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 2 tan^{2} (x)

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

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

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

用替代法求解

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

令

: tan (x) = u

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

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

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

- 1 \cdot (1 - u^{2}) + \frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

- 1 \cdot (1 - u^{2}) + \frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

- 1 \cdot (1 - u^{2}) : - (1 - u^{2})

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

乘以:

1 \cdot (1 - u^{2}) = (1 - u^{2})

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

化简

\frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2}) : 2 u^{2}

\frac{2 u ^{2}}{1 - u ^{2}} (1 - u^{2})

分式相乘:

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

= \frac{2 u ^{2} ( 1 - u ^{2} )}{1 - u ^{2}}

约分:

1 - u^{2}

= 2 u^{2}

化简

0 \cdot (1 - u^{2}) : 0

0 \cdot (1 - u^{2})

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

展开

- (1 - u^{2}) + 2 u^{2} : - 1 + 3 u^{2}

- (1 - u^{2}) + 2 u^{2}

- (1 - u^{2}) : - 1 + u^{2}

- (1 - u^{2})

打开括号

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

使用加减运算法则

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

= - 1 + u^{2}

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

同类项相加:

u^{2} + 2 u^{2} = 3 u^{2}

= - 1 + 3 u^{2}

- 1 + 3 u^{2} = 0

将

1

到右边

- 1 + 3 u^{2} = 0

两边加上

1

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

化简

3 u^{2} = 1

3 u^{2} = 1

两边除以

3

3 u^{2} = 1

两边除以

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

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

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

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

u = tan (x)

代回

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

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

使用反三角函数性质

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

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

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

使用反三角函数性质

tan (x) = - \frac{1}{3}

tan (x) = - \frac{1}{3}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

合并所有解

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

以小数形式表示解

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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