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受欢迎的三角函数 >

-1=(tan(x)-(3/4))/(1+tan(x)(3/4))

解答

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

解答

x = - 0.14189 \dots + πn

+1

度数

x = - 8.13010 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

交换两边

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

用替代法求解

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

令

: tan (x) = u

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

\frac{u - \frac{3}{4}}{1 + u \frac{3}{4}} = - 1 : u = - \frac{1}{7}

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

化简

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

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

化简

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

1 + u \frac{3}{4}

乘

u \frac{3}{4} : \frac{3 u}{4}

u \frac{3}{4}

分式相乘:

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

= \frac{3 u}{4}

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

将项转换为分式:

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

= \frac{1 \cdot 4}{4} + \frac{3 u}{4}

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{1 \cdot 4 + 3 u}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 + 3 u}{4}

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

化简

u - \frac{3}{4} : \frac{4 u - 3}{4}

u - \frac{3}{4}

将项转换为分式:

u = \frac{u 4}{4}

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

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{u \cdot 4 - 3}{4}

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

分式相除:

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

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

约分:

4

= \frac{u \cdot 4 - 3}{4 + 3 u}

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

在两边乘以

4 + 3 u

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

在两边乘以

4 + 3 u

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

化简

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

化简

\frac{4 u - 3}{4 + 3 u} (4 + 3 u) : 4 u - 3

\frac{4 u - 3}{4 + 3 u} (4 + 3 u)

分式相乘:

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

= \frac{( 4 u - 3 ) ( 4 + 3 u )}{4 + 3 u}

约分:

4 + 3 u

= 4 u - 3

化简

- 1 \cdot (4 + 3 u) : - (4 + 3 u)

- 1 \cdot (4 + 3 u)

乘以:

1 \cdot (4 + 3 u) = (4 + 3 u)

= - (3 u + 4)

4 u - 3 = - (4 + 3 u)

4 u - 3 = - (4 + 3 u)

4 u - 3 = - (4 + 3 u)

展开

- (4 + 3 u) : - 4 - 3 u

- (4 + 3 u)

打开括号

= - (4) - (3 u)

使用加减运算法则

+ (- a) = - a

= - 4 - 3 u

4 u - 3 = - 4 - 3 u

将

3

到右边

4 u - 3 = - 4 - 3 u

两边加上

3

4 u - 3 + 3 = - 4 - 3 u + 3

化简

4 u = - 3 u - 1

4 u = - 3 u - 1

将

3 u

para o lado esquerdo

4 u = - 3 u - 1

两边加上

3 u

4 u + 3 u = - 3 u - 1 + 3 u

化简

7 u = - 1

7 u = - 1

两边除以

7

7 u = - 1

两边除以

7

\frac{7 u}{7} = \frac{- 1}{7}

化简

u = - \frac{1}{7}

u = - \frac{1}{7}

验证解

找到无定义的点（奇点）:

u = - \frac{4}{3}

取

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

的分母，令其等于零

解

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

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

将

1

到右边

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

两边减去

1

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

化简

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

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

在两边乘以

4

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

在两边乘以

4

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

化简

3 u = - 4

3 u = - 4

两边除以

3

3 u = - 4

两边除以

3

\frac{3 u}{3} = \frac{- 4}{3}

化简

u = - \frac{4}{3}

u = - \frac{4}{3}

以下点无定义

u = - \frac{4}{3}

将不在定义域的点与解相综合:

u = - \frac{1}{7}

u = tan (x)

代回

tan (x) = - \frac{1}{7}

tan (x) = - \frac{1}{7}

tan (x) = - \frac{1}{7} : x = arctan (- \frac{1}{7}) + πn

tan (x) = - \frac{1}{7}

使用反三角函数性质

tan (x) = - \frac{1}{7}

tan (x) = - \frac{1}{7}

的通解

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

x = arctan (- \frac{1}{7}) + πn

x = arctan (- \frac{1}{7}) + πn

合并所有解

x = arctan (- \frac{1}{7}) + πn

以小数形式表示解

x = - 0.14189 \dots + πn

作图

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