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受欢迎的三角函数 >

sec(2x)+tan(2x)=8

解答

sec (2 x) + tan (2 x) = 8

解答

x = \frac{1.32208 \dots}{2} + πn

+1

度数

x = 37.87498 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec (2 x) + tan (2 x) = 8

两边减去

8

sec (2 x) + tan (2 x) - 8 = 0

用 sin, cos 表示

\frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 8 = 0

化简

\frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 8 : \frac{1 + sin ( 2 x ) - 8 cos ( 2 x )}{cos ( 2 x )}

\frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} - 8

合并分式

\frac{1}{cos ( 2 x )} + \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{1 + sin ( 2 x )}{cos ( 2 x )}

使用法则

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

= \frac{1 + sin ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - 8

将项转换为分式:

8 = \frac{8 cos ( 2 x )}{cos ( 2 x )}

= \frac{1 + sin ( 2 x )}{cos ( 2 x )} - \frac{8 cos ( 2 x )}{cos ( 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 + sin ( 2 x ) - 8 cos ( 2 x )}{cos ( 2 x )}

\frac{1 + sin ( 2 x ) - 8 cos ( 2 x )}{cos ( 2 x )} = 0

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

1 + sin (2 x) - 8 cos (2 x) = 0

两边加上

8 cos (2 x)

1 + sin (2 x) = 8 cos (2 x)

两边进行平方

(1 + sin (2 x))^{2} = (8 cos (2 x))^{2}

两边减去

(8 cos (2 x))^{2}

(1 + sin (2 x))^{2} - 64 cos^{2} (2 x) = 0

使用三角恒等式改写

(1 + sin (2 x))^{2} - 64 cos^{2} (2 x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= (1 + sin (2 x))^{2} - 64 (1 - sin^{2} (2 x))

化简

(1 + sin (2 x))^{2} - 64 (1 - sin^{2} (2 x)) : 65 sin^{2} (2 x) + 2 sin (2 x) - 63

(1 + sin (2 x))^{2} - 64 (1 - sin^{2} (2 x))

(1 + sin (2 x))^{2} : 1 + 2 sin (2 x) + sin^{2} (2 x)

使用完全平方公式:

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

a = 1, b = sin (2 x)

= 1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

化简

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x) : 1 + 2 sin (2 x) + sin^{2} (2 x)

1^{2} + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (2 x) + sin^{2} (2 x)

数字相乘:

2 \cdot 1 = 2

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

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

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 64 (1 - sin^{2} (2 x))

乘开

- 64 (1 - sin^{2} (2 x)) : - 64 + 64 sin^{2} (2 x)

- 64 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 64, b = 1, c = sin^{2} (2 x)

= - 64 \cdot 1 - (- 64) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 64 \cdot 1 + 64 sin^{2} (2 x)

数字相乘:

64 \cdot 1 = 64

= - 64 + 64 sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 64 + 64 sin^{2} (2 x)

化简

1 + 2 sin (2 x) + sin^{2} (2 x) - 64 + 64 sin^{2} (2 x) : 65 sin^{2} (2 x) + 2 sin (2 x) - 63

1 + 2 sin (2 x) + sin^{2} (2 x) - 64 + 64 sin^{2} (2 x)

对同类项分组

= 2 sin (2 x) + sin^{2} (2 x) + 64 sin^{2} (2 x) + 1 - 64

同类项相加:

sin^{2} (2 x) + 64 sin^{2} (2 x) = 65 sin^{2} (2 x)

= 2 sin (2 x) + 65 sin^{2} (2 x) + 1 - 64

数字相加/相减:

1 - 64 = - 63

= 65 sin^{2} (2 x) + 2 sin (2 x) - 63

= 65 sin^{2} (2 x) + 2 sin (2 x) - 63

= 65 sin^{2} (2 x) + 2 sin (2 x) - 63

- 63 + 2 sin (2 x) + 65 sin^{2} (2 x) = 0

用替代法求解

- 63 + 2 sin (2 x) + 65 sin^{2} (2 x) = 0

令

: sin (2 x) = u

- 63 + 2 u + 65 u^{2} = 0

- 63 + 2 u + 65 u^{2} = 0 : u = \frac{63}{65}, u = - 1

- 63 + 2 u + 65 u^{2} = 0

改写成标准形式

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

65 u^{2} + 2 u - 63 = 0

使用求根公式求解

65 u^{2} + 2 u - 63 = 0

二次方程求根公式:

若

a = 65, b = 2, c = - 63

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 65 ( - 63 )}{2 \cdot 65}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 65 ( - 63 )}{2 \cdot 65}

2^{2} - 4 \cdot 65 (- 63) = 128

2^{2} - 4 \cdot 65 (- 63)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 65 \cdot 63

数字相乘:

4 \cdot 65 \cdot 63 = 16380

= 2^{2} + 16380

2^{2} = 4

= 4 + 16380

数字相加:

4 + 16380 = 16384

= 16384

因式分解数字:

16384 = 12 8^{2}

= 12 8^{2}

使用根式运算法则:

n a^{n} = a

12 8^{2} = 128

= 128

u_{1, 2} = \frac{- 2 \pm 128}{2 \cdot 65}

将解分隔开

u_{1} = \frac{- 2 + 128}{2 \cdot 65}, u_{2} = \frac{- 2 - 128}{2 \cdot 65}

u = \frac{- 2 + 128}{2 \cdot 65} : \frac{63}{65}

\frac{- 2 + 128}{2 \cdot 65}

数字相加/相减:

- 2 + 128 = 126

= \frac{126}{2 \cdot 65}

数字相乘:

2 \cdot 65 = 130

= \frac{126}{130}

约分:

2

= \frac{63}{65}

u = \frac{- 2 - 128}{2 \cdot 65} : - 1

\frac{- 2 - 128}{2 \cdot 65}

数字相减:

- 2 - 128 = - 130

= \frac{- 130}{2 \cdot 65}

数字相乘:

2 \cdot 65 = 130

= \frac{- 130}{130}

使用分式法则:

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

= - \frac{130}{130}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{63}{65}, u = - 1

u = sin (2 x)

代回

sin (2 x) = \frac{63}{65}, sin (2 x) = - 1

sin (2 x) = \frac{63}{65}, sin (2 x) = - 1

sin (2 x) = \frac{63}{65} : x = \frac{arcsin ( \frac{63}{65} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

sin (2 x) = \frac{63}{65}

使用反三角函数性质

sin (2 x) = \frac{63}{65}

sin (2 x) = \frac{63}{65}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 x = arcsin (\frac{63}{65}) + 2 πn, 2 x = π - arcsin (\frac{63}{65}) + 2 πn

2 x = arcsin (\frac{63}{65}) + 2 πn, 2 x = π - arcsin (\frac{63}{65}) + 2 πn

解

2 x = arcsin (\frac{63}{65}) + 2 πn : x = \frac{arcsin ( \frac{63}{65} )}{2} + πn

2 x = arcsin (\frac{63}{65}) + 2 πn

两边除以

2

2 x = arcsin (\frac{63}{65}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{63}{65} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{63}{65} )}{2} + πn

x = \frac{arcsin ( \frac{63}{65} )}{2} + πn

解

2 x = π - arcsin (\frac{63}{65}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

2 x = π - arcsin (\frac{63}{65}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{63}{65}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

x = \frac{arcsin ( \frac{63}{65} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

sin (2 x) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

解

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

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

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

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

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = \frac{arcsin ( \frac{63}{65} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn, x = \frac{3 π}{4} + πn

将解代入原方程进行验证

将它们代入

sec (2 x) + tan (2 x) = 8

检验解是否符合

去除与方程不符的解。

检验

\frac{arcsin ( \frac{63}{65} )}{2} + πn

的解:

真

\frac{arcsin ( \frac{63}{65} )}{2} + πn

代入

n = 1

\frac{arcsin ( \frac{63}{65} )}{2} + π 1

对于

sec (2 x) + tan (2 x) = 8

代入

x = \frac{arcsin ( \frac{63}{65} )}{2} + π 1

sec (2 (\frac{arcsin ( \frac{63}{65} )}{2} + π 1)) + tan (2 (\frac{arcsin ( \frac{63}{65} )}{2} + π 1)) = 8

整理后得

8 = 8

\Rightarrow 真

检验

\frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

的解:

假

\frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + πn

代入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + π 1

对于

sec (2 x) + tan (2 x) = 8

代入

x = \frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + π 1

sec (2 (\frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + π 1)) + tan (2 (\frac{π}{2} - \frac{arcsin ( \frac{63}{65} )}{2} + π 1)) = 8

整理后得

- 8 = 8

\Rightarrow 假

检验

\frac{3 π}{4} + πn

的解:

假

\frac{3 π}{4} + πn

代入

n = 1

\frac{3 π}{4} + π 1

对于

sec (2 x) + tan (2 x) = 8

代入

x = \frac{3 π}{4} + π 1

sec (2 (\frac{3 π}{4} + π 1)) + tan (2 (\frac{3 π}{4} + π 1)) = 8

未定义

\Rightarrow 假

x = \frac{arcsin ( \frac{63}{65} )}{2} + πn

以小数形式表示解

x = \frac{1.32208 \dots}{2} + πn

作图

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