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受欢迎的三角函数 >

tan(3x)=5tan(x)

解答

tan (3 x) = 5 tan (x)

解答

x = πn, x = - 0.36136 \dots + πn, x = 0.36136 \dots + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = - 20.70481 \dots^{\circ} + 18 0^{\circ} n, x = 20.70481 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (3 x) = 5 tan (x)

两边减去

5 tan (x)

tan (3 x) - 5 tan (x) = 0

使用三角恒等式改写

tan (3 x) - 5 tan (x)

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

tan (3 x)

使用三角恒等式改写

tan (3 x)

改写为

= tan (2 x + x)

使用角和恒等式:

tan (s + t) = \frac{tan ( s ) + tan ( t )}{1 - tan ( s ) tan ( t )}

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

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

使用倍角公式:

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

= \frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

化简

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )} : \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} + tan ( x )}{1 - \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )} tan ( x )}

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

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

分式相乘:

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

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

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

2 tan (x) tan (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= 2 tan^{2} (x)

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

= \frac{\frac{2 t a n ( x )}{- t a n ^{2} ( x ) + 1} + tan ( x )}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

化简

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

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

将项转换为分式:

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

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

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 ( x ) + tan ( x ) ( 1 - tan ^{2} ( x ) )}{1 - tan ^{2} ( x )}

乘开

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

2 tan (x) + tan (x) (1 - tan^{2} (x))

乘开

tan (x) (1 - tan^{2} (x)) : tan (x) - tan^{3} (x)

tan (x) (1 - tan^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = tan (x), b = 1, c = tan^{2} (x)

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

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

化简

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

1 tan (x) - tan^{2} (x) tan (x)

1 \cdot tan (x) = tan (x)

1 tan (x)

乘以:

1 \cdot tan (x) = tan (x)

= tan (x)

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

tan^{2} (x) tan (x)

使用指数法则:

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

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

= tan^{2 + 1} (x)

数字相加:

2 + 1 = 3

= tan^{3} (x)

= tan (x) - tan^{3} (x)

= tan (x) - tan^{3} (x)

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

同类项相加:

2 tan (x) + tan (x) = 3 tan (x)

= 3 tan (x) - tan^{3} (x)

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

= \frac{\frac{3 t a n ( x ) - t a n ^{3} ( x )}{1 - t a n ^{2} ( x )}}{1 - \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

使用分式法则:

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{( 1 - tan ^{2} ( x ) ) ( 1 - \frac{2 t a n ^{2} ( x )}{1 - t a n ^{2} ( x )} )}

化简

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

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

将项转换为分式:

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

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

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 ( 1 - tan ^{2} ( x ) ) - 2 tan ^{2} ( x )}{1 - tan ^{2} ( x )}

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

1 (1 - tan^{2} (x)) - 2 tan^{2} (x)

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

1 (1 - tan^{2} (x))

乘以:

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

= 1 - tan^{2} (x)

去除括号:

(a) = a

= 1 - tan^{2} (x)

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

同类项相加:

- tan^{2} (x) - 2 tan^{2} (x) = - 3 tan^{2} (x)

= 1 - 3 tan^{2} (x)

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{\frac{- 3 t a n ^{2} ( x ) + 1}{- t a n ^{2} ( x ) + 1} ( - tan ^{2} ( x ) + 1 )}

乘

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

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

分式相乘:

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

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

约分:

1 - tan^{2} (x)

= 1 - 3 tan^{2} (x)

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

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

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x)

化简

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x) : \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - 5 tan (x)

将项转换为分式:

5 tan (x) = \frac{5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

= \frac{3 tan ( x ) - tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )} - \frac{5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

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{3 tan ( x ) - tan ^{3} ( x ) - 5 tan ( x ) ( 1 - 3 tan ^{2} ( x ) )}{1 - 3 tan ^{2} ( x )}

乘开

3 tan (x) - tan^{3} (x) - 5 tan (x) (1 - 3 tan^{2} (x)) : - 2 tan (x) + 14 tan^{3} (x)

3 tan (x) - tan^{3} (x) - 5 tan (x) (1 - 3 tan^{2} (x))

乘开

- 5 tan (x) (1 - 3 tan^{2} (x)) : - 5 tan (x) + 15 tan^{3} (x)

- 5 tan (x) (1 - 3 tan^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 5 tan (x), b = 1, c = 3 tan^{2} (x)

= - 5 tan (x) \cdot 1 - (- 5 tan (x)) \cdot 3 tan^{2} (x)

使用加减运算法则

- (- a) = a

= - 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x)

化简

- 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x) : - 5 tan (x) + 15 tan^{3} (x)

- 5 \cdot 1 \cdot tan (x) + 5 \cdot 3 tan^{2} (x) tan (x)

5 \cdot 1 \cdot tan (x) = 5 tan (x)

5 \cdot 1 \cdot tan (x)

数字相乘:

5 \cdot 1 = 5

= 5 tan (x)

5 \cdot 3 tan^{2} (x) tan (x) = 15 tan^{3} (x)

5 \cdot 3 tan^{2} (x) tan (x)

数字相乘:

5 \cdot 3 = 15

= 15 tan^{2} (x) tan (x)

使用指数法则:

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

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

= 15 tan^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 15 tan^{3} (x)

= - 5 tan (x) + 15 tan^{3} (x)

= - 5 tan (x) + 15 tan^{3} (x)

= 3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x)

化简

3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x) : - 2 tan (x) + 14 tan^{3} (x)

3 tan (x) - tan^{3} (x) - 5 tan (x) + 15 tan^{3} (x)

同类项相加:

- tan^{3} (x) + 15 tan^{3} (x) = 14 tan^{3} (x)

= 3 tan (x) + 14 tan^{3} (x) - 5 tan (x)

同类项相加:

3 tan (x) - 5 tan (x) = - 2 tan (x)

= - 2 tan (x) + 14 tan^{3} (x)

= - 2 tan (x) + 14 tan^{3} (x)

= \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

= \frac{- 2 tan ( x ) + 14 tan ^{3} ( x )}{1 - 3 tan ^{2} ( x )}

\frac{14 tan ^{3} ( x ) - 2 tan ( x )}{1 - 3 tan ^{2} ( x )} = 0

用替代法求解

\frac{14 tan ^{3} ( x ) - 2 tan ( x )}{1 - 3 tan ^{2} ( x )} = 0

令

: tan (x) = u

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0 : u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}} = 0

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

14 u^{3} - 2 u = 0

解

14 u^{3} - 2 u = 0 : u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

14 u^{3} - 2 u = 0

因式分解

14 u^{3} - 2 u : 2 u (7 u + 1) (7 u - 1)

14 u^{3} - 2 u

因式分解出通项

2 u : 2 u (7 u^{2} - 1)

14 u^{3} - 2 u

使用指数法则:

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

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

= 14 u^{2} u - 2 u

将

14

改写为

2 \cdot 7

= 2 \cdot 7 u^{2} u - 2 u

因式分解出通项

2 u

= 2 u (7 u^{2} - 1)

= 2 u (7 u^{2} - 1)

分解

7 u^{2} - 1 : (7 u + 1) (7 u - 1)

7 u^{2} - 1

将

7 u^{2} - 1

改写为

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

7 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

7 = (7)^{2}

= (7)^{2} u^{2} - 1

将

1

改写为

1^{2}

= (7)^{2} u^{2} - 1^{2}

使用指数法则:

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

(7)^{2} u^{2} = (7 u)^{2}

= (7 u)^{2} - 1^{2}

= (7 u)^{2} - 1^{2}

使用平方差公式:

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

(7 u)^{2} - 1^{2} = (7 u + 1) (7 u - 1)

= (7 u + 1) (7 u - 1)

= 2 u (7 u + 1) (7 u - 1)

2 u (7 u + 1) (7 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 7 u + 1 = 0 or 7 u - 1 = 0

解

7 u + 1 = 0 : u = - \frac{7}{7}

7 u + 1 = 0

将

1

到右边

7 u + 1 = 0

两边减去

1

7 u + 1 - 1 = 0 - 1

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

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

化简

\frac{7 u}{7} : u

\frac{7 u}{7}

约分:

7

= u

化简

\frac{- 1}{7} : - \frac{7}{7}

\frac{- 1}{7}

使用分式法则:

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

= - \frac{1}{7}

- \frac{1}{7}

有理化:

- \frac{7}{7}

- \frac{1}{7}

乘以共轭根式

\frac{7}{7}

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

1 \cdot 7 = 7

77 = 7

77

使用根式运算法则:

a a = a

77 = 7

= 7

= - \frac{7}{7}

= - \frac{7}{7}

u = - \frac{7}{7}

u = - \frac{7}{7}

u = - \frac{7}{7}

解

7 u - 1 = 0 : u = \frac{7}{7}

7 u - 1 = 0

将

1

到右边

7 u - 1 = 0

两边加上

1

7 u - 1 + 1 = 0 + 1

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

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

化简

\frac{7 u}{7} : u

\frac{7 u}{7}

约分:

7

= u

化简

\frac{1}{7} : \frac{7}{7}

\frac{1}{7}

乘以共轭根式

\frac{7}{7}

= \frac{1 \cdot 7}{7 7}

1 \cdot 7 = 7

77 = 7

77

使用根式运算法则:

a a = a

77 = 7

= 7

= \frac{7}{7}

u = \frac{7}{7}

u = \frac{7}{7}

u = \frac{7}{7}

解为

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

验证解

找到无定义的点（奇点）:

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

取

\frac{14 u ^{3} - 2 u}{1 - 3 u ^{2}}

的分母，令其等于零

解

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

1 - 3 u^{2} = 0

将

1

到右边

1 - 3 u^{2} = 0

两边减去

1

1 - 3 u^{2} - 1 = 0 - 1

化简

- 3 u^{2} = - 1

- 3 u^{2} = - 1

两边除以

- 3

- 3 u^{2} = - 1

两边除以

- 3

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

化简

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

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

对于

x^{2} = f (a)

解为

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

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

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

\frac{1}{3}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{3}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{3}

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

- \frac{1}{3}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{3}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{3}

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

以下点无定义

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

将不在定义域的点与解相综合:

u = 0, u = - \frac{7}{7}, u = \frac{7}{7}

u = tan (x)

代回

tan (x) = 0, tan (x) = - \frac{7}{7}, tan (x) = \frac{7}{7}

tan (x) = 0, tan (x) = - \frac{7}{7}, tan (x) = \frac{7}{7}

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 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}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

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

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

使用反三角函数性质

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

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

的通解

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

x = arctan (- \frac{7}{7}) + πn

x = arctan (- \frac{7}{7}) + πn

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

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

使用反三角函数性质

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

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

的通解

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

x = arctan (\frac{7}{7}) + πn

x = arctan (\frac{7}{7}) + πn

合并所有解

x = πn, x = arctan (- \frac{7}{7}) + πn, x = arctan (\frac{7}{7}) + πn

以小数形式表示解

x = πn, x = - 0.36136 \dots + πn, x = 0.36136 \dots + πn

作图

流行的例子

2cos(2x)-sqrt(2)=0

2 cos (2 x) - 2 = 0

tan(x+20)*tan(x-20)=1

tan (x + 2 0^{\circ}) \cdot tan (x - 2 0^{\circ}) = 1

18cos(10x)+2=10

18 cos (10 x) + 2 = 10

cos(2θ)=sin(2θ)

cos (2 θ) = sin (2 θ)

2cos(x/2)=sqrt(3)

2 cos (\frac{x}{2}) = 3