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受欢迎的三角函数 >

(2tan(3x))/(1-tan^2(3x))=1

解答

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

解答

x = - \frac{1.17809 \dots}{3} + \frac{πn}{3}, x = \frac{0.39269 \dots}{3} + \frac{πn}{3}

+1

度数

x = - 22. 5^{\circ} + 6 0^{\circ} n, x = 7. 5^{\circ} + 6 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: tan (3 x) = u

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

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

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

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

化简

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

化简

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

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

分式相乘:

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

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

约分:

1 - u^{2}

= 2 u

化简

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

1 \cdot (1 - u^{2})

乘以:

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

= (1 - u^{2})

去除括号:

(a) = a

= 1 - u^{2}

2 u = 1 - u^{2}

2 u = 1 - u^{2}

2 u = 1 - u^{2}

解

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

2 u = 1 - u^{2}

交换两边

1 - u^{2} = 2 u

将

2 u

para o lado esquerdo

1 - u^{2} = 2 u

两边减去

2 u

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

化简

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

(- 2)^{2} - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2}{- 2}

使用分式法则:

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

= - \frac{2 + 2 2}{2}

消掉

\frac{2 + 2 2}{2} : 1 + 2

\frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

打开括号

= - (1) - (2)

使用加减运算法则

+ (- a) = - a

= - 1 - 2

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

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2}{- 2}

使用分式法则:

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

2 - 22 = - (22 - 2)

= \frac{2 2 - 2}{2}

分解

22 - 2 : 2 (2 - 1)

22 - 2

改写为

= 22 - 2 \cdot 1

因式分解出通项

2

= 2 (2 - 1)

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

数字相除:

\frac{2}{2} = 1

= 2 - 1

二次方程组的解是:

u = - 1 - 2, u = 2 - 1

u = - 1 - 2, u = 2 - 1

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

u = tan (3 x)

代回

tan (3 x) = - 1 - 2, tan (3 x) = 2 - 1

tan (3 x) = - 1 - 2, tan (3 x) = 2 - 1

tan (3 x) = - 1 - 2 : x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

tan (3 x) = - 1 - 2

使用反三角函数性质

tan (3 x) = - 1 - 2

tan (3 x) = - 1 - 2

的通解

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

3 x = arctan (- 1 - 2) + πn

3 x = arctan (- 1 - 2) + πn

解

3 x = arctan (- 1 - 2) + πn : x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

3 x = arctan (- 1 - 2) + πn

化简

arctan (- 1 - 2) + πn : - arctan (1 + 2) + πn

arctan (- 1 - 2) + πn

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- 1 - 2) = - arctan (1 + 2)

= - arctan (1 + 2) + πn

3 x = - arctan (1 + 2) + πn

两边除以

3

3 x = - arctan (1 + 2) + πn

两边除以

3

\frac{3 x}{3} = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

化简

x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}

tan (3 x) = 2 - 1 : x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

tan (3 x) = 2 - 1

使用反三角函数性质

tan (3 x) = 2 - 1

tan (3 x) = 2 - 1

的通解

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

3 x = arctan (2 - 1) + πn

3 x = arctan (2 - 1) + πn

解

3 x = arctan (2 - 1) + πn : x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

3 x = arctan (2 - 1) + πn

两边除以

3

3 x = arctan (2 - 1) + πn

两边除以

3

\frac{3 x}{3} = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

化简

x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

合并所有解

x = - \frac{arctan ( 1 + 2 )}{3} + \frac{πn}{3}, x = \frac{arctan ( 2 - 1 )}{3} + \frac{πn}{3}

以小数形式表示解

x = - \frac{1.17809 \dots}{3} + \frac{πn}{3}, x = \frac{0.39269 \dots}{3} + \frac{πn}{3}

作图

流行的例子

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