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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

- 3

tan (\frac{x}{2}) tan (x) + 3 = 0

令

: u = \frac{x}{2}

tan (u) tan (2 u) + 3 = 0

使用三角恒等式改写

3 + tan (2 u) tan (u)

使用倍角公式:

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

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

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

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

分式相乘:

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

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

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

2 tan (u) tan (u)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= 2 tan^{2} (u)

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

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

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

用替代法求解

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

令

: tan (u) = u

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

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

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

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

化简

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

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

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

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

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

解

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

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

乘开

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

3 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

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

数字相乘:

3 \cdot 1 = 3

= 3 - 3 u^{2}

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

同类项相加:

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

3 - u^{2} = 0

将

3

到右边

3 - u^{2} = 0

两边减去

3

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

化简

- u^{2} = - 3

- u^{2} = - 3

两边除以

- 1

- u^{2} = - 3

两边除以

- 1

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

化简

u^{2} = 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

u = 3, u = - 3

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

3 + \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 = 3, u = - 3

u = tan (u)

代回

tan (u) = 3, tan (u) = - 3

tan (u) = 3, tan (u) = - 3

tan (u) = 3 : u = \frac{π}{3} + πn

tan (u) = 3

tan (u) = 3

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}

u = \frac{π}{3} + πn

u = \frac{π}{3} + πn

tan (u) = - 3 : u = \frac{2 π}{3} + πn

tan (u) = - 3

tan (u) = - 3

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}

u = \frac{2 π}{3} + πn

u = \frac{2 π}{3} + πn

合并所有解

u = \frac{π}{3} + πn, u = \frac{2 π}{3} + πn

u = \frac{x}{2}

代回

\frac{x}{2} = \frac{π}{3} + πn : x = \frac{2 π}{3} + 2 πn

\frac{x}{2} = \frac{π}{3} + πn

在两边乘以

2

\frac{x}{2} = \frac{π}{3} + πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{3} + 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{3} + 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{3} + 2 πn : \frac{2 π}{3} + 2 πn

2 \cdot \frac{π}{3} + 2 πn

乘

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

分式相乘:

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

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

\frac{x}{2} = \frac{2 π}{3} + πn : x = \frac{4 π}{3} + 2 πn

\frac{x}{2} = \frac{2 π}{3} + πn

在两边乘以

2

\frac{x}{2} = \frac{2 π}{3} + πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{2 π}{3} + 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{2 π}{3} + 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{2 π}{3} + 2 πn : \frac{4 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

分式相乘:

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

= \frac{2 π 2}{3}

数字相乘:

2 \cdot 2 = 4

= \frac{4 π}{3}

= \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

作图

流行的例子

solvefor x,h=tan(x-pi/6)

sin(x/3)cos(6x)-cos(x/3)sin(6x)=0