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受欢迎的三角函数 >

tan(6x)-3tan(3x)=0

解答

tan (6 x) - 3 tan (3 x) = 0

解答

x = \frac{πn}{3}, x = \frac{5 π + 6 πn}{18}, x = \frac{π + 6 πn}{18}

+1

度数

x = 0^{\circ} + 6 0^{\circ} n, x = 5 0^{\circ} + 6 0^{\circ} n, x = 1 0^{\circ} + 6 0^{\circ} n

求解步骤

tan (6 x) - 3 tan (3 x) = 0

令

: u = 3 x

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

使用三角恒等式改写

tan (2 u) - 3 tan (u)

使用倍角公式:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘开

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 3 tan (u), b = 1, c = tan^{2} (u)

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

使用加减运算法则

- (- a) = a

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

化简

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

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

3 \cdot 1 \cdot tan (u) = 3 tan (u)

3 \cdot 1 \cdot tan (u)

数字相乘:

3 \cdot 1 = 3

= 3 tan (u)

3 tan^{2} (u) tan (u) = 3 tan^{3} (u)

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 3 tan^{3} (u)

= - 3 tan (u) + 3 tan^{3} (u)

= - 3 tan (u) + 3 tan^{3} (u)

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

同类项相加:

2 tan (u) - 3 tan (u) = - tan (u)

= - tan (u) + 3 tan^{3} (u)

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

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

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

用替代法求解

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

令

: tan (u) = u

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

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

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

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

- u + 3 u^{3} = 0

解

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

- u + 3 u^{3} = 0

因式分解

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

- u + 3 u^{3}

因式分解出通项

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

3 u^{3} - u

使用指数法则:

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

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

= 3 u^{2} u - u

因式分解出通项

u

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

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

分解

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

3 u^{2} - 1

将

3 u^{2} - 1

改写为

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

3 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(3)^{2} u^{2} = (3 u)^{2}

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (3 u + 1) (3 u - 1)

= u (3 u + 1) (3 u - 1)

u (3 u + 1) (3 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 3 u + 1 = 0 or 3 u - 1 = 0

解

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

3 u + 1 = 0

将

1

到右边

3 u + 1 = 0

两边减去

1

3 u + 1 - 1 = 0 - 1

化简

3 u = - 1

3 u = - 1

两边除以

3

3 u = - 1

两边除以

3

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

化简

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

化简

\frac{3 u}{3} : u

\frac{3 u}{3}

约分:

3

= u

化简

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

使用分式法则:

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

= - \frac{1}{3}

- \frac{1}{3}

有理化:

- \frac{3}{3}

- \frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

解

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

3 u - 1 = 0

将

1

到右边

3 u - 1 = 0

两边加上

1

3 u - 1 + 1 = 0 + 1

化简

3 u = 1

3 u = 1

两边除以

3

3 u = 1

两边除以

3

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

化简

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

化简

\frac{3 u}{3} : u

\frac{3 u}{3}

约分:

3

= u

化简

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

解为

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

u = tan (u)

代回

tan (u) = 0, tan (u) = - \frac{3}{3}, tan (u) = \frac{3}{3}

tan (u) = 0, tan (u) = - \frac{3}{3}, tan (u) = \frac{3}{3}

tan (u) = 0 : u = πn

tan (u) = 0

tan (u) = 0

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 = 0 + πn

u = 0 + πn

解

u = 0 + πn : u = πn

u = 0 + πn

0 + πn = πn

u = πn

u = πn

tan (u) = - \frac{3}{3} : u = \frac{5 π}{6} + πn

tan (u) = - \frac{3}{3}

tan (u) = - \frac{3}{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{5 π}{6} + πn

u = \frac{5 π}{6} + πn

tan (u) = \frac{3}{3} : u = \frac{π}{6} + πn

tan (u) = \frac{3}{3}

tan (u) = \frac{3}{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}{6} + πn

u = \frac{π}{6} + πn

合并所有解

u = πn, u = \frac{5 π}{6} + πn, u = \frac{π}{6} + πn

u = 3 x

代回

3 x = πn : x = \frac{πn}{3}

3 x = πn

两边除以

3

3 x = πn

两边除以

3

\frac{3 x}{3} = \frac{πn}{3}

化简

x = \frac{πn}{3}

x = \frac{πn}{3}

3 x = \frac{5 π}{6} + πn : x = \frac{5 π + 6 πn}{18}

3 x = \frac{5 π}{6} + πn

两边除以

3

3 x = \frac{5 π}{6} + πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{5 π}{6}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{5 π}{6}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{5 π}{6}}{3} + \frac{πn}{3} : \frac{5 π + 6 πn}{18}

\frac{\frac{5 π}{6}}{3} + \frac{πn}{3}

使用法则

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

= \frac{\frac{5 π}{6} + πn}{3}

化简

\frac{5 π}{6} + πn : \frac{5 π + 6 πn}{6}

\frac{5 π}{6} + πn

将项转换为分式:

πn = \frac{πn 6}{6}

= \frac{5 π}{6} + \frac{πn \cdot 6}{6}

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{5 π + πn \cdot 6}{6}

= \frac{\frac{5 π + 6 πn}{6}}{3}

使用分式法则:

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

= \frac{5 π + πn \cdot 6}{6 \cdot 3}

数字相乘:

6 \cdot 3 = 18

= \frac{5 π + 6 πn}{18}

x = \frac{5 π + 6 πn}{18}

x = \frac{5 π + 6 πn}{18}

x = \frac{5 π + 6 πn}{18}

3 x = \frac{π}{6} + πn : x = \frac{π + 6 πn}{18}

3 x = \frac{π}{6} + πn

两边除以

3

3 x = \frac{π}{6} + πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{π}{6}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{π}{6}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{π}{6}}{3} + \frac{πn}{3} : \frac{π + 6 πn}{18}

\frac{\frac{π}{6}}{3} + \frac{πn}{3}

使用法则

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

= \frac{\frac{π}{6} + πn}{3}

化简

\frac{π}{6} + πn : \frac{π + 6 πn}{6}

\frac{π}{6} + πn

将项转换为分式:

πn = \frac{πn 6}{6}

= \frac{π}{6} + \frac{πn \cdot 6}{6}

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{π + πn \cdot 6}{6}

= \frac{\frac{π + 6 πn}{6}}{3}

使用分式法则:

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

= \frac{π + πn \cdot 6}{6 \cdot 3}

数字相乘:

6 \cdot 3 = 18

= \frac{π + 6 πn}{18}

x = \frac{π + 6 πn}{18}

x = \frac{π + 6 πn}{18}

x = \frac{π + 6 πn}{18}

x = \frac{πn}{3}, x = \frac{5 π + 6 πn}{18}, x = \frac{π + 6 πn}{18}

作图

流行的例子

sin (a) = \frac{2}{5}

cos^2(θ)-cos(θ)-1=0

cos^{2} (θ) - cos (θ) - 1 = 0

3cot^2(x)-4csc(x)=1

3 cot^{2} (x) - 4 csc (x) = 1

cos(x)=sin(2x)+cos(3x)

cos (x) = sin (2 x) + cos (3 x)

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

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