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受欢迎的三角函数 >

1/(sec(2x)-1)-1/(sec(2x)+1)=6

解答

\frac{1}{sec ( 2 x ) - 1} - \frac{1}{sec ( 2 x ) + 1} = 6

解答

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{5 π}{12} + πn, x = \frac{7 π}{12} + πn

+1

度数

x = 1 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n, x = 10 5^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1}{sec ( 2 x ) - 1} - \frac{1}{sec ( 2 x ) + 1} = 6

用替代法求解

\frac{1}{sec ( 2 x ) - 1} - \frac{1}{sec ( 2 x ) + 1} = 6

令

: sec (2 x) = u

\frac{1}{u - 1} - \frac{1}{u + 1} = 6

\frac{1}{u - 1} - \frac{1}{u + 1} = 6 : u = \frac{2 3}{3}, u = - \frac{2 3}{3}

\frac{1}{u - 1} - \frac{1}{u + 1} = 6

乘以最小公倍数

\frac{1}{u - 1} - \frac{1}{u + 1} = 6

找到

u - 1, u + 1

的最小公倍数:

(u - 1) (u + 1)

u - 1, u + 1

最小公倍数 (LCM)

计算出由出现在

u - 1

或

u + 1

中的因子组成的表达式

= (u - 1) (u + 1)

乘以最小公倍数=

(u - 1) (u + 1)

\frac{1}{u - 1} (u - 1) (u + 1) - \frac{1}{u + 1} (u - 1) (u + 1) = 6 (u - 1) (u + 1)

化简

\frac{1}{u - 1} (u - 1) (u + 1) - \frac{1}{u + 1} (u - 1) (u + 1) = 6 (u - 1) (u + 1)

化简

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

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

分式相乘:

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

= \frac{1 \cdot ( u - 1 ) ( u + 1 )}{u - 1}

约分:

u - 1

= 1 \cdot (u + 1)

整理后得

= u + 1

化简

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

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

分式相乘:

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

= - \frac{1 \cdot ( u - 1 ) ( u + 1 )}{u + 1}

约分:

u + 1

= - 1 \cdot (u - 1)

乘以:

1 \cdot (u - 1) = (u - 1)

= - (u - 1)

u + 1 - (u - 1) = 6 (u - 1) (u + 1)

u + 1 - (u - 1) = 6 (u - 1) (u + 1)

u + 1 - (u - 1) = 6 (u - 1) (u + 1)

解

u + 1 - (u - 1) = 6 (u - 1) (u + 1) : u = \frac{2 3}{3}, u = - \frac{2 3}{3}

u + 1 - (u - 1) = 6 (u - 1) (u + 1)

展开

u + 1 - (u - 1) : 2

u + 1 - (u - 1)

- (u - 1) : - u + 1

- (u - 1)

打开括号

= - (u) - (- 1)

使用加减运算法则

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

= - u + 1

= u + 1 - u + 1

化简

u + 1 - u + 1 : 2

u + 1 - u + 1

对同类项分组

= u - u + 1 + 1

同类项相加:

u - u = 0

= 1 + 1

数字相加:

1 + 1 = 2

= 2

= 2

展开

6 (u - 1) (u + 1) : 6 u^{2} - 6

6 (u - 1) (u + 1)

乘开

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

(u - 1) (u + 1)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = u, b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= u^{2} - 1

= 6 (u^{2} - 1)

乘开

6 (u^{2} - 1) : 6 u^{2} - 6

6 (u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = 6, b = u^{2}, c = 1

= 6 u^{2} - 6 \cdot 1

数字相乘:

6 \cdot 1 = 6

= 6 u^{2} - 6

= 6 u^{2} - 6

2 = 6 u^{2} - 6

交换两边

6 u^{2} - 6 = 2

将

6

到右边

6 u^{2} - 6 = 2

两边加上

6

6 u^{2} - 6 + 6 = 2 + 6

化简

6 u^{2} = 8

6 u^{2} = 8

两边除以

6

6 u^{2} = 8

两边除以

6

\frac{6 u ^{2}}{6} = \frac{8}{6}

化简

u^{2} = \frac{4}{3}

u^{2} = \frac{4}{3}

对于

x^{2} = f (a)

解为

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

u = \frac{4}{3}, u = - \frac{4}{3}

\frac{4}{3} = \frac{2 3}{3}

\frac{4}{3}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

\frac{2}{3}

有理化:

\frac{2 3}{3}

\frac{2}{3}

乘以共轭根式

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

= \frac{2 3}{3}

- \frac{4}{3} = - \frac{2 3}{3}

- \frac{4}{3}

化简

\frac{4}{3} : \frac{2}{3}

\frac{4}{3}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

= - \frac{2}{3}

- \frac{2}{3}

有理化:

- \frac{2 3}{3}

- \frac{2}{3}

乘以共轭根式

\frac{3}{3}

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

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

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

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

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

\frac{1}{u - 1} - \frac{1}{u + 1}

的分母，令其等于零

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

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

u = sec (2 x)

代回

sec (2 x) = \frac{2 3}{3}, sec (2 x) = - \frac{2 3}{3}

sec (2 x) = \frac{2 3}{3}, sec (2 x) = - \frac{2 3}{3}

sec (2 x) = \frac{2 3}{3} : x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

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

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

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

解

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

2 x = \frac{11 π}{6} + 2 πn

两边除以

2

2 x = \frac{11 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

使用分式法则:

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

= \frac{11 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

sec (2 x) = - \frac{2 3}{3} : x = \frac{5 π}{12} + πn, x = \frac{7 π}{12} + πn

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

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

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 x = \frac{5 π}{6} + 2 πn, 2 x = \frac{7 π}{6} + 2 πn

2 x = \frac{5 π}{6} + 2 πn, 2 x = \frac{7 π}{6} + 2 πn

解

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

解

2 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{12} + πn

2 x = \frac{7 π}{6} + 2 πn

两边除以

2

2 x = \frac{7 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

使用分式法则:

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

= \frac{7 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{5 π}{12} + πn, x = \frac{7 π}{12} + πn

合并所有解

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{5 π}{12} + πn, x = \frac{7 π}{12} + πn

作图

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