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受欢迎的三角函数 >

6sec(2x)+3tan(2x)-9=0

解答

6 sec (2 x) + 3 tan (2 x) - 9 = 0

解答

x = \frac{0.56432 \dots}{2} + πn, x = - \frac{1.20782 \dots}{2} + πn

+1

度数

x = 16.16676 \dots^{\circ} + 18 0^{\circ} n, x = - 34.60171 \dots^{\circ} + 18 0^{\circ} n

求解步骤

6 sec (2 x) + 3 tan (2 x) - 9 = 0

用 sin, cos 表示

6 \cdot \frac{1}{cos ( 2 x )} + 3 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 9 = 0

化简

6 \cdot \frac{1}{cos ( 2 x )} + 3 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 9 : \frac{6 + 3 sin ( 2 x ) - 9 cos ( 2 x )}{cos ( 2 x )}

6 \cdot \frac{1}{cos ( 2 x )} + 3 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 9

6 \cdot \frac{1}{cos ( 2 x )} = \frac{6}{cos ( 2 x )}

6 \cdot \frac{1}{cos ( 2 x )}

分式相乘:

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

= \frac{1 \cdot 6}{cos ( 2 x )}

数字相乘:

1 \cdot 6 = 6

= \frac{6}{cos ( 2 x )}

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

3 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

分式相乘:

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

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

= \frac{6}{cos ( 2 x )} + \frac{3 sin ( 2 x )}{cos ( 2 x )} - 9

合并分式

\frac{6}{cos ( 2 x )} + \frac{3 sin ( 2 x )}{cos ( 2 x )} : \frac{6 + 3 sin ( 2 x )}{cos ( 2 x )}

使用法则

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

= \frac{6 + 3 sin ( 2 x )}{cos ( 2 x )}

= \frac{3 sin ( 2 x ) + 6}{cos ( 2 x )} - 9

将项转换为分式:

9 = \frac{9 cos ( 2 x )}{cos ( 2 x )}

= \frac{6 + sin ( 2 x ) \cdot 3}{cos ( 2 x )} - \frac{9 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{6 + sin ( 2 x ) \cdot 3 - 9 cos ( 2 x )}{cos ( 2 x )}

\frac{6 + 3 sin ( 2 x ) - 9 cos ( 2 x )}{cos ( 2 x )} = 0

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

6 + 3 sin (2 x) - 9 cos (2 x) = 0

两边加上

9 cos (2 x)

6 + 3 sin (2 x) = 9 cos (2 x)

两边进行平方

(6 + 3 sin (2 x))^{2} = (9 cos (2 x))^{2}

两边减去

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

(6 + 3 sin (2 x))^{2} - 81 cos^{2} (2 x) = 0

使用三角恒等式改写

(6 + 3 sin (2 x))^{2} - 81 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)

= (6 + 3 sin (2 x))^{2} - 81 (1 - sin^{2} (2 x))

化简

(6 + 3 sin (2 x))^{2} - 81 (1 - sin^{2} (2 x)) : 90 sin^{2} (2 x) + 36 sin (2 x) - 45

(6 + 3 sin (2 x))^{2} - 81 (1 - sin^{2} (2 x))

(6 + 3 sin (2 x))^{2} : 36 + 36 sin (2 x) + 9 sin^{2} (2 x)

使用完全平方公式:

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

a = 6, b = 3 sin (2 x)

= 6^{2} + 2 \cdot 6 \cdot 3 sin (2 x) + (3 sin (2 x))^{2}

化简

6^{2} + 2 \cdot 6 \cdot 3 sin (2 x) + (3 sin (2 x))^{2} : 36 + 36 sin (2 x) + 9 sin^{2} (2 x)

6^{2} + 2 \cdot 6 \cdot 3 sin (2 x) + (3 sin (2 x))^{2}

6^{2} = 36

6^{2}

6^{2} = 36

= 36

2 \cdot 6 \cdot 3 sin (2 x) = 36 sin (2 x)

2 \cdot 6 \cdot 3 sin (2 x)

数字相乘:

2 \cdot 6 \cdot 3 = 36

= 36 sin (2 x)

(3 sin (2 x))^{2} = 9 sin^{2} (2 x)

(3 sin (2 x))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

3^{2} = 9

= 9 sin^{2} (2 x)

= 36 + 36 sin (2 x) + 9 sin^{2} (2 x)

= 36 + 36 sin (2 x) + 9 sin^{2} (2 x)

= 36 + 36 sin (2 x) + 9 sin^{2} (2 x) - 81 (1 - sin^{2} (2 x))

乘开

- 81 (1 - sin^{2} (2 x)) : - 81 + 81 sin^{2} (2 x)

- 81 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 81, b = 1, c = sin^{2} (2 x)

= - 81 \cdot 1 - (- 81) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 81 \cdot 1 + 81 sin^{2} (2 x)

数字相乘:

81 \cdot 1 = 81

= - 81 + 81 sin^{2} (2 x)

= 36 + 36 sin (2 x) + 9 sin^{2} (2 x) - 81 + 81 sin^{2} (2 x)

化简

36 + 36 sin (2 x) + 9 sin^{2} (2 x) - 81 + 81 sin^{2} (2 x) : 90 sin^{2} (2 x) + 36 sin (2 x) - 45

36 + 36 sin (2 x) + 9 sin^{2} (2 x) - 81 + 81 sin^{2} (2 x)

对同类项分组

= 36 sin (2 x) + 9 sin^{2} (2 x) + 81 sin^{2} (2 x) + 36 - 81

同类项相加:

9 sin^{2} (2 x) + 81 sin^{2} (2 x) = 90 sin^{2} (2 x)

= 36 sin (2 x) + 90 sin^{2} (2 x) + 36 - 81

数字相加/相减:

36 - 81 = - 45

= 90 sin^{2} (2 x) + 36 sin (2 x) - 45

= 90 sin^{2} (2 x) + 36 sin (2 x) - 45

= 90 sin^{2} (2 x) + 36 sin (2 x) - 45

- 45 + 36 sin (2 x) + 90 sin^{2} (2 x) = 0

用替代法求解

- 45 + 36 sin (2 x) + 90 sin^{2} (2 x) = 0

令

: sin (2 x) = u

- 45 + 36 u + 90 u^{2} = 0

- 45 + 36 u + 90 u^{2} = 0 : u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

- 45 + 36 u + 90 u^{2} = 0

改写成标准形式

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

90 u^{2} + 36 u - 45 = 0

使用求根公式求解

90 u^{2} + 36 u - 45 = 0

二次方程求根公式:

若

a = 90, b = 36, c = - 45

u_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 90 ( - 45 )}{2 \cdot 90}

u_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 90 ( - 45 )}{2 \cdot 90}

3 6^{2} - 4 \cdot 90 (- 45) = 546

3 6^{2} - 4 \cdot 90 (- 45)

使用法则

- (- a) = a

= 3 6^{2} + 4 \cdot 90 \cdot 45

数字相乘:

4 \cdot 90 \cdot 45 = 16200

= 3 6^{2} + 16200

3 6^{2} = 1296

= 1296 + 16200

数字相加:

1296 + 16200 = 17496

= 17496

17496

质因数分解:

2^{3} \cdot 3^{7}

17496

17496

除以

217496 = 8748 \cdot 2

= 2 \cdot 8748

8748

除以

28748 = 4374 \cdot 2

= 2 \cdot 2 \cdot 4374

4374

除以

24374 = 2187 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2187

2187

除以

32187 = 729 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 729

729

除以

3729 = 243 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 243

243

除以

3243 = 81 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 81

81

除以

381 = 27 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 27

27

除以

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{7}

= 3^{7} \cdot 2^{3}

使用指数法则:

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

= 3^{6} \cdot 2^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 3^{6} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{6} 2 \cdot 3

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

3^{6} = 3^{\frac{6}{2}} = 3^{3}

= 3^{3} \cdot 2 2 \cdot 3

整理后得

= 546

u_{1, 2} = \frac{- 36 \pm 54 6}{2 \cdot 90}

将解分隔开

u_{1} = \frac{- 36 + 54 6}{2 \cdot 90}, u_{2} = \frac{- 36 - 54 6}{2 \cdot 90}

u = \frac{- 36 + 54 6}{2 \cdot 90} : \frac{- 2 + 3 6}{10}

\frac{- 36 + 54 6}{2 \cdot 90}

数字相乘:

2 \cdot 90 = 180

= \frac{- 36 + 54 6}{180}

分解

- 36 + 546 : 18 (- 2 + 36)

- 36 + 546

改写为

= - 18 \cdot 2 + 18 \cdot 36

因式分解出通项

18

= 18 (- 2 + 36)

= \frac{18 ( - 2 + 3 6 )}{180}

约分:

18

= \frac{- 2 + 3 6}{10}

u = \frac{- 36 - 54 6}{2 \cdot 90} : - \frac{2 + 3 6}{10}

\frac{- 36 - 54 6}{2 \cdot 90}

数字相乘:

2 \cdot 90 = 180

= \frac{- 36 - 54 6}{180}

分解

- 36 - 546 : - 18 (2 + 36)

- 36 - 546

改写为

= - 18 \cdot 2 - 18 \cdot 36

因式分解出通项

18

= - 18 (2 + 36)

= - \frac{18 ( 2 + 3 6 )}{180}

约分:

18

= - \frac{2 + 3 6}{10}

二次方程组的解是:

u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

u = sin (2 x)

代回

sin (2 x) = \frac{- 2 + 3 6}{10}, sin (2 x) = - \frac{2 + 3 6}{10}

sin (2 x) = \frac{- 2 + 3 6}{10}, sin (2 x) = - \frac{2 + 3 6}{10}

sin (2 x) = \frac{- 2 + 3 6}{10} : x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

sin (2 x) = \frac{- 2 + 3 6}{10}

使用反三角函数性质

sin (2 x) = \frac{- 2 + 3 6}{10}

sin (2 x) = \frac{- 2 + 3 6}{10}

的通解

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

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

解

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn : x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

两边除以

2

2 x = arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

解

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

sin (2 x) = - \frac{2 + 3 6}{10} : x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

sin (2 x) = - \frac{2 + 3 6}{10}

使用反三角函数性质

sin (2 x) = - \frac{2 + 3 6}{10}

sin (2 x) = - \frac{2 + 3 6}{10}

的通解

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

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn, 2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn, 2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

解

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn : x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

2 x = arcsin (- \frac{2 + 3 6}{10}) + 2 πn

化简

arcsin (- \frac{2 + 3 6}{10}) + 2 πn : - arcsin (\frac{2 + 3 6}{10}) + 2 πn

arcsin (- \frac{2 + 3 6}{10}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{2 + 3 6}{10}) = - arcsin (\frac{2 + 3 6}{10})

= - arcsin (\frac{2 + 3 6}{10}) + 2 πn

2 x = - arcsin (\frac{2 + 3 6}{10}) + 2 πn

两边除以

2

2 x = - arcsin (\frac{2 + 3 6}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

解

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

两边除以

2

2 x = π + arcsin (\frac{2 + 3 6}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

合并所有解

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn, x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

将解代入原方程进行验证

将它们代入

6 sec (2 x) + 3 tan (2 x) - 9 = 0

检验解是否符合

去除与方程不符的解。

检验

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

的解:

真

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

代入

n = 1

\frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

对于

6 sec (2 x) + 3 tan (2 x) - 9 = 0

代入

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

6 sec 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + 3 tan 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 9 = 0

整理后得

0 = 0

\Rightarrow 真

检验

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

的解:

假

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn

代入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

对于

6 sec (2 x) + 3 tan (2 x) - 9 = 0

代入

x = \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1

6 sec 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + 3 tan 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 9 = 0

整理后得

- 18 = 0

\Rightarrow 假

检验

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

的解:

真

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

代入

n = 1

- \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

对于

6 sec (2 x) + 3 tan (2 x) - 9 = 0

代入

x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

6 sec 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + 3 tan 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 9 = 0

整理后得

0 = 0

\Rightarrow 真

检验

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

的解:

假

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

代入

n = 1

\frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

对于

6 sec (2 x) + 3 tan (2 x) - 9 = 0

代入

x = \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1

6 sec 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + 3 tan 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 9 = 0

整理后得

- 18 = 0

\Rightarrow 假

x = \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + πn, x = - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + πn

以小数形式表示解

x = \frac{0.56432 \dots}{2} + πn, x = - \frac{1.20782 \dots}{2} + πn

作图

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