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受欢迎的三角函数 >

1/(tan(x))-2tan(x)=-1/4

解答

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

解答

x = - 0.57451 \dots + πn, x = 0.65766 \dots + πn

+1

度数

x = - 32.91754 \dots^{\circ} + 18 0^{\circ} n, x = 37.68118 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

用替代法求解

\frac{1}{tan ( x )} - 2 tan (x) = - \frac{1}{4}

令

: tan (x) = u

\frac{1}{u} - 2 u = - \frac{1}{4}

\frac{1}{u} - 2 u = - \frac{1}{4} : u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

\frac{1}{u} - 2 u = - \frac{1}{4}

乘以最小公倍数

\frac{1}{u} - 2 u = - \frac{1}{4}

找到

u, 4

的最小公倍数:

4 u

u, 4

最小公倍数 (LCM)

计算出由出现在

u

或

4

中的因子组成的表达式

= 4 u

乘以最小公倍数=

4 u

\frac{1}{u} \cdot 4 u - 2 u \cdot 4 u = - \frac{1}{4} \cdot 4 u

化简

\frac{1}{u} \cdot 4 u - 2 u \cdot 4 u = - \frac{1}{4} \cdot 4 u

化简

\frac{1}{u} \cdot 4 u : 4

\frac{1}{u} \cdot 4 u

分式相乘:

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

= \frac{1 \cdot 4 u}{u}

约分:

u

= 1 \cdot 4

数字相乘:

1 \cdot 4 = 4

= 4

化简

- 2 u \cdot 4 u : - 8 u^{2}

- 2 u \cdot 4 u

数字相乘:

2 \cdot 4 = 8

= - 8 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 8 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 8 u^{2}

化简

- \frac{1}{4} \cdot 4 u : - u

- \frac{1}{4} \cdot 4 u

分式相乘:

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

= - \frac{1 \cdot 4}{4} u

约分:

4

= - u \cdot 1

乘以:

u \cdot 1 = u

= - u

4 - 8 u^{2} = - u

4 - 8 u^{2} = - u

4 - 8 u^{2} = - u

解

4 - 8 u^{2} = - u : u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

4 - 8 u^{2} = - u

将

u

para o lado esquerdo

4 - 8 u^{2} = - u

两边加上

u

4 - 8 u^{2} + u = - u + u

化简

4 - 8 u^{2} + u = 0

4 - 8 u^{2} + u = 0

改写成标准形式

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

- 8 u^{2} + u + 4 = 0

使用求根公式求解

- 8 u^{2} + u + 4 = 0

二次方程求根公式:

若

a = - 8, b = 1, c = 4

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

1^{2} - 4 (- 8) \cdot 4 = 129

1^{2} - 4 (- 8) \cdot 4

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 8) \cdot 4

使用法则

- (- a) = a

= 1 + 4 \cdot 8 \cdot 4

数字相乘:

4 \cdot 8 \cdot 4 = 128

= 1 + 128

数字相加:

1 + 128 = 129

= 129

u_{1, 2} = \frac{- 1 \pm 129}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 1 + 129}{2 ( - 8 )}, u_{2} = \frac{- 1 - 129}{2 ( - 8 )}

u = \frac{- 1 + 129}{2 ( - 8 )} : - \frac{- 1 + 129}{16}

\frac{- 1 + 129}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 1 + 129}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 1 + 129}{- 16}

使用分式法则:

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

= - \frac{- 1 + 129}{16}

u = \frac{- 1 - 129}{2 ( - 8 )} : \frac{1 + 129}{16}

\frac{- 1 - 129}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 1 - 129}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 1 - 129}{- 16}

使用分式法则:

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

- 1 - 129 = - (1 + 129)

= \frac{1 + 129}{16}

二次方程组的解是:

u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

u = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

验证解

找到无定义的点（奇点）:

u = 0

取

\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 = - \frac{- 1 + 129}{16}, u = \frac{1 + 129}{16}

u = tan (x)

代回

tan (x) = - \frac{- 1 + 129}{16}, tan (x) = \frac{1 + 129}{16}

tan (x) = - \frac{- 1 + 129}{16}, tan (x) = \frac{1 + 129}{16}

tan (x) = - \frac{- 1 + 129}{16} : x = arctan (- \frac{- 1 + 129}{16}) + πn

tan (x) = - \frac{- 1 + 129}{16}

使用反三角函数性质

tan (x) = - \frac{- 1 + 129}{16}

tan (x) = - \frac{- 1 + 129}{16}

的通解

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

x = arctan (- \frac{- 1 + 129}{16}) + πn

x = arctan (- \frac{- 1 + 129}{16}) + πn

tan (x) = \frac{1 + 129}{16} : x = arctan (\frac{1 + 129}{16}) + πn

tan (x) = \frac{1 + 129}{16}

使用反三角函数性质

tan (x) = \frac{1 + 129}{16}

tan (x) = \frac{1 + 129}{16}

的通解

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

x = arctan (\frac{1 + 129}{16}) + πn

x = arctan (\frac{1 + 129}{16}) + πn

合并所有解

x = arctan (- \frac{- 1 + 129}{16}) + πn, x = arctan (\frac{1 + 129}{16}) + πn

以小数形式表示解

x = - 0.57451 \dots + πn, x = 0.65766 \dots + πn

作图

流行的例子

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tan(4x+20)*cot(x+50)=1

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