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受欢迎的三角函数 >

2/(tan(x))=3-tan(x)

解答

\frac{2}{tan ( x )} = 3 - tan (x)

解答

x = \frac{π}{4} + πn, x = 1.10714 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{2}{tan ( x )} = 3 - tan (x)

用替代法求解

\frac{2}{tan ( x )} = 3 - tan (x)

令

: tan (x) = u

\frac{2}{u} = 3 - u

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

\frac{2}{u} = 3 - u

在两边乘以

u

\frac{2}{u} = 3 - u

在两边乘以

u

\frac{2}{u} u = 3 u - uu

化简

- uu : - u^{2}

\frac{2}{u} u = 3 u - 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}

2 = 3 u - u^{2}

2 = 3 u - u^{2}

解

2 = 3 u - u^{2} : u = 1, u = 2

2 = 3 u - u^{2}

交换两边

3 u - u^{2} = 2

将

2

para o lado esquerdo

3 u - u^{2} = 2

两边减去

2

3 u - u^{2} - 2 = 2 - 2

化简

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

3^{2} - 4 (- 1) (- 2) = 1

3^{2} - 4 (- 1) (- 2)

使用法则

- (- a) = a

= 3^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 ( - 1 )}

将解分隔开

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

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

\frac{- 3 + 1}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 1}

数字相加/相减:

- 3 + 1 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

\frac{- 3 - 1}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 1}

数字相减:

- 3 - 1 = - 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

使用分式法则:

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

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

二次方程组的解是:

u = 1, u = 2

u = 1, u = 2

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{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 = 1, u = 2

u = tan (x)

代回

tan (x) = 1, tan (x) = 2

tan (x) = 1, tan (x) = 2

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

使用反三角函数性质

tan (x) = 2

tan (x) = 2

的通解

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

x = arctan (2) + πn

x = arctan (2) + πn

合并所有解

x = \frac{π}{4} + πn, x = arctan (2) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = 1.10714 \dots + πn

作图

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tan(x)= 1/(5^{1/2)}

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sqrt(3)csc(x/2)-2=0

3 csc (\frac{x}{2}) - 2 = 0