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受欢迎的三角函数 >

10sec(2x)+5tan(2x)-15=0

解答

10 sec (2 x) + 5 tan (2 x) - 15 = 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

10 sec (2 x) + 5 tan (2 x) - 15 = 0

用 sin, cos 表示

10 \cdot \frac{1}{cos ( 2 x )} + 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 15 = 0

化简

10 \cdot \frac{1}{cos ( 2 x )} + 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 15 : \frac{10 + 5 sin ( 2 x ) - 15 cos ( 2 x )}{cos ( 2 x )}

10 \cdot \frac{1}{cos ( 2 x )} + 5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} - 15

10 \cdot \frac{1}{cos ( 2 x )} = \frac{10}{cos ( 2 x )}

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

分式相乘:

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

= \frac{1 \cdot 10}{cos ( 2 x )}

数字相乘:

1 \cdot 10 = 10

= \frac{10}{cos ( 2 x )}

5 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} = \frac{5 sin ( 2 x )}{cos ( 2 x )}

5 \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 5}{cos ( 2 x )}

= \frac{10}{cos ( 2 x )} + \frac{5 sin ( 2 x )}{cos ( 2 x )} - 15

合并分式

\frac{10}{cos ( 2 x )} + \frac{5 sin ( 2 x )}{cos ( 2 x )} : \frac{10 + 5 sin ( 2 x )}{cos ( 2 x )}

使用法则

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

= \frac{10 + 5 sin ( 2 x )}{cos ( 2 x )}

= \frac{5 sin ( 2 x ) + 10}{cos ( 2 x )} - 15

将项转换为分式:

15 = \frac{15 cos ( 2 x )}{cos ( 2 x )}

= \frac{10 + sin ( 2 x ) \cdot 5}{cos ( 2 x )} - \frac{15 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{10 + sin ( 2 x ) \cdot 5 - 15 cos ( 2 x )}{cos ( 2 x )}

\frac{10 + 5 sin ( 2 x ) - 15 cos ( 2 x )}{cos ( 2 x )} = 0

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

10 + 5 sin (2 x) - 15 cos (2 x) = 0

两边加上

15 cos (2 x)

10 + 5 sin (2 x) = 15 cos (2 x)

两边进行平方

(10 + 5 sin (2 x))^{2} = (15 cos (2 x))^{2}

两边减去

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

(10 + 5 sin (2 x))^{2} - 225 cos^{2} (2 x) = 0

使用三角恒等式改写

(10 + 5 sin (2 x))^{2} - 225 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)

= (10 + 5 sin (2 x))^{2} - 225 (1 - sin^{2} (2 x))

化简

(10 + 5 sin (2 x))^{2} - 225 (1 - sin^{2} (2 x)) : 250 sin^{2} (2 x) + 100 sin (2 x) - 125

(10 + 5 sin (2 x))^{2} - 225 (1 - sin^{2} (2 x))

(10 + 5 sin (2 x))^{2} : 100 + 100 sin (2 x) + 25 sin^{2} (2 x)

使用完全平方公式:

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

a = 10, b = 5 sin (2 x)

= 1 0^{2} + 2 \cdot 10 \cdot 5 sin (2 x) + (5 sin (2 x))^{2}

化简

1 0^{2} + 2 \cdot 10 \cdot 5 sin (2 x) + (5 sin (2 x))^{2} : 100 + 100 sin (2 x) + 25 sin^{2} (2 x)

1 0^{2} + 2 \cdot 10 \cdot 5 sin (2 x) + (5 sin (2 x))^{2}

1 0^{2} = 100

1 0^{2}

1 0^{2} = 100

= 100

2 \cdot 10 \cdot 5 sin (2 x) = 100 sin (2 x)

2 \cdot 10 \cdot 5 sin (2 x)

数字相乘:

2 \cdot 10 \cdot 5 = 100

= 100 sin (2 x)

(5 sin (2 x))^{2} = 25 sin^{2} (2 x)

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

使用指数法则:

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

= 5^{2} sin^{2} (2 x)

5^{2} = 25

= 25 sin^{2} (2 x)

= 100 + 100 sin (2 x) + 25 sin^{2} (2 x)

= 100 + 100 sin (2 x) + 25 sin^{2} (2 x)

= 100 + 100 sin (2 x) + 25 sin^{2} (2 x) - 225 (1 - sin^{2} (2 x))

乘开

- 225 (1 - sin^{2} (2 x)) : - 225 + 225 sin^{2} (2 x)

- 225 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 225, b = 1, c = sin^{2} (2 x)

= - 225 \cdot 1 - (- 225) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 225 \cdot 1 + 225 sin^{2} (2 x)

数字相乘:

225 \cdot 1 = 225

= - 225 + 225 sin^{2} (2 x)

= 100 + 100 sin (2 x) + 25 sin^{2} (2 x) - 225 + 225 sin^{2} (2 x)

化简

100 + 100 sin (2 x) + 25 sin^{2} (2 x) - 225 + 225 sin^{2} (2 x) : 250 sin^{2} (2 x) + 100 sin (2 x) - 125

100 + 100 sin (2 x) + 25 sin^{2} (2 x) - 225 + 225 sin^{2} (2 x)

对同类项分组

= 100 sin (2 x) + 25 sin^{2} (2 x) + 225 sin^{2} (2 x) + 100 - 225

同类项相加:

25 sin^{2} (2 x) + 225 sin^{2} (2 x) = 250 sin^{2} (2 x)

= 100 sin (2 x) + 250 sin^{2} (2 x) + 100 - 225

数字相加/相减:

100 - 225 = - 125

= 250 sin^{2} (2 x) + 100 sin (2 x) - 125

= 250 sin^{2} (2 x) + 100 sin (2 x) - 125

= 250 sin^{2} (2 x) + 100 sin (2 x) - 125

- 125 + 100 sin (2 x) + 250 sin^{2} (2 x) = 0

用替代法求解

- 125 + 100 sin (2 x) + 250 sin^{2} (2 x) = 0

令

: sin (2 x) = u

- 125 + 100 u + 250 u^{2} = 0

- 125 + 100 u + 250 u^{2} = 0 : u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

- 125 + 100 u + 250 u^{2} = 0

改写成标准形式

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

250 u^{2} + 100 u - 125 = 0

使用求根公式求解

250 u^{2} + 100 u - 125 = 0

二次方程求根公式:

若

a = 250, b = 100, c = - 125

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 250 ( - 125 )}{2 \cdot 250}

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 250 ( - 125 )}{2 \cdot 250}

10 0^{2} - 4 \cdot 250 (- 125) = 1506

10 0^{2} - 4 \cdot 250 (- 125)

使用法则

- (- a) = a

= 10 0^{2} + 4 \cdot 250 \cdot 125

数字相乘:

4 \cdot 250 \cdot 125 = 125000

= 10 0^{2} + 125000

10 0^{2} = 10000

= 10000 + 125000

数字相加:

10000 + 125000 = 135000

= 135000

135000

质因数分解:

2^{3} \cdot 3^{3} \cdot 5^{4}

135000

= 5^{4} \cdot 2^{3} \cdot 3^{3}

使用指数法则:

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

= 5^{4} \cdot 2^{2} \cdot 3^{2} \cdot 2 \cdot 3

使用根式运算法则:

= 2^{2} 3^{2} 5^{4} 2 \cdot 3

使用根式运算法则:

2^{2} = 2

= 2 3^{2} 5^{4} 2 \cdot 3

使用根式运算法则:

3^{2} = 3

= 2 \cdot 3 5^{4} 2 \cdot 3

使用根式运算法则:

5^{4} = 5^{\frac{4}{2}} = 5^{2}

= 5^{2} \cdot 2 \cdot 3 2 \cdot 3

整理后得

= 1506

u_{1, 2} = \frac{- 100 \pm 150 6}{2 \cdot 250}

将解分隔开

u_{1} = \frac{- 100 + 150 6}{2 \cdot 250}, u_{2} = \frac{- 100 - 150 6}{2 \cdot 250}

u = \frac{- 100 + 150 6}{2 \cdot 250} : \frac{- 2 + 3 6}{10}

\frac{- 100 + 150 6}{2 \cdot 250}

数字相乘:

2 \cdot 250 = 500

= \frac{- 100 + 150 6}{500}

分解

- 100 + 1506 : 50 (- 2 + 36)

- 100 + 1506

改写为

= - 50 \cdot 2 + 50 \cdot 36

因式分解出通项

50

= 50 (- 2 + 36)

= \frac{50 ( - 2 + 3 6 )}{500}

约分:

50

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

u = \frac{- 100 - 150 6}{2 \cdot 250} : - \frac{2 + 3 6}{10}

\frac{- 100 - 150 6}{2 \cdot 250}

数字相乘:

2 \cdot 250 = 500

= \frac{- 100 - 150 6}{500}

分解

- 100 - 1506 : - 50 (2 + 36)

- 100 - 1506

改写为

= - 50 \cdot 2 - 50 \cdot 36

因式分解出通项

50

= - 50 (2 + 36)

= - \frac{50 ( 2 + 3 6 )}{500}

约分:

50

= - \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

10 sec (2 x) + 5 tan (2 x) - 15 = 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

10 sec (2 x) + 5 tan (2 x) - 15 = 0

代入

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

10 sec 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + 5 tan 2 \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 15 = 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

10 sec (2 x) + 5 tan (2 x) - 15 = 0

代入

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

10 sec 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 + 5 tan 2 \frac{π}{2} - \frac{arcsin ( \frac{- 2 + 3 6}{10} )}{2} + π 1 - 15 = 0

整理后得

- 30 = 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

10 sec (2 x) + 5 tan (2 x) - 15 = 0

代入

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

10 sec 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + 5 tan 2 - \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 15 = 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

10 sec (2 x) + 5 tan (2 x) - 15 = 0

代入

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

10 sec 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 + 5 tan 2 \frac{π}{2} + \frac{arcsin ( \frac{2 + 3 6}{10} )}{2} + π 1 - 15 = 0

整理后得

- 30 = 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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