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受欢迎的三角函数 >

tan(2x)+sec(2x)=4

解答

tan (2 x) + sec (2 x) = 4

解答

x = \frac{1.08083 \dots}{2} + πn

+1

度数

x = 30.96375 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 x) + sec (2 x) = 4

两边减去

4

tan (2 x) + sec (2 x) - 4 = 0

用 sin, cos 表示

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4 = 0

化简

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4 : \frac{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x )}{cos ( 2 x )} + \frac{1}{cos ( 2 x )} - 4

合并分式

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

使用法则

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

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

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - 4

将项转换为分式:

4 = \frac{4 cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) + 1}{cos ( 2 x )} - \frac{4 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{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x ) + 1 - 4 cos ( 2 x )}{cos ( 2 x )} = 0

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

sin (2 x) + 1 - 4 cos (2 x) = 0

两边加上

4 cos (2 x)

sin (2 x) + 1 = 4 cos (2 x)

两边进行平方

(sin (2 x) + 1)^{2} = (4 cos (2 x))^{2}

两边减去

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

(sin (2 x) + 1)^{2} - 16 cos^{2} (2 x) = 0

使用三角恒等式改写

(1 + sin (2 x))^{2} - 16 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} - 16 (1 - sin^{2} (2 x))

化简

(1 + sin (2 x))^{2} - 16 (1 - sin^{2} (2 x)) : 17 sin^{2} (2 x) + 2 sin (2 x) - 15

(1 + sin (2 x))^{2} - 16 (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) - 16 (1 - sin^{2} (2 x))

乘开

- 16 (1 - sin^{2} (2 x)) : - 16 + 16 sin^{2} (2 x)

- 16 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (2 x)

= - 16 \cdot 1 - (- 16) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (2 x)

数字相乘:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (2 x)

= 1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x)

化简

1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x) : 17 sin^{2} (2 x) + 2 sin (2 x) - 15

1 + 2 sin (2 x) + sin^{2} (2 x) - 16 + 16 sin^{2} (2 x)

对同类项分组

= 2 sin (2 x) + sin^{2} (2 x) + 16 sin^{2} (2 x) + 1 - 16

同类项相加:

sin^{2} (2 x) + 16 sin^{2} (2 x) = 17 sin^{2} (2 x)

= 2 sin (2 x) + 17 sin^{2} (2 x) + 1 - 16

数字相加/相减:

1 - 16 = - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

= 17 sin^{2} (2 x) + 2 sin (2 x) - 15

- 15 + 17 sin^{2} (2 x) + 2 sin (2 x) = 0

用替代法求解

- 15 + 17 sin^{2} (2 x) + 2 sin (2 x) = 0

令

: sin (2 x) = u

- 15 + 17 u^{2} + 2 u = 0

- 15 + 17 u^{2} + 2 u = 0 : u = \frac{15}{17}, u = - 1

- 15 + 17 u^{2} + 2 u = 0

改写成标准形式

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

17 u^{2} + 2 u - 15 = 0

使用求根公式求解

17 u^{2} + 2 u - 15 = 0

二次方程求根公式:

若

a = 17, b = 2, c = - 15

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 17 ( - 15 )}{2 \cdot 17}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 17 ( - 15 )}{2 \cdot 17}

2^{2} - 4 \cdot 17 (- 15) = 32

2^{2} - 4 \cdot 17 (- 15)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 17 \cdot 15

数字相乘:

4 \cdot 17 \cdot 15 = 1020

= 2^{2} + 1020

2^{2} = 4

= 4 + 1020

数字相加:

4 + 1020 = 1024

= 1024

因式分解数字:

1024 = 3 2^{2}

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

3 2^{2} = 32

= 32

u_{1, 2} = \frac{- 2 \pm 32}{2 \cdot 17}

将解分隔开

u_{1} = \frac{- 2 + 32}{2 \cdot 17}, u_{2} = \frac{- 2 - 32}{2 \cdot 17}

u = \frac{- 2 + 32}{2 \cdot 17} : \frac{15}{17}

\frac{- 2 + 32}{2 \cdot 17}

数字相加/相减:

- 2 + 32 = 30

= \frac{30}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{30}{34}

约分:

2

= \frac{15}{17}

u = \frac{- 2 - 32}{2 \cdot 17} : - 1

\frac{- 2 - 32}{2 \cdot 17}

数字相减:

- 2 - 32 = - 34

= \frac{- 34}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{- 34}{34}

使用分式法则:

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

= - \frac{34}{34}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{15}{17}, u = - 1

u = sin (2 x)

代回

sin (2 x) = \frac{15}{17}, sin (2 x) = - 1

sin (2 x) = \frac{15}{17}, sin (2 x) = - 1

sin (2 x) = \frac{15}{17} : x = \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

sin (2 x) = \frac{15}{17}

使用反三角函数性质

sin (2 x) = \frac{15}{17}

sin (2 x) = \frac{15}{17}

的通解

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

2 x = arcsin (\frac{15}{17}) + 2 πn, 2 x = π - arcsin (\frac{15}{17}) + 2 πn

2 x = arcsin (\frac{15}{17}) + 2 πn, 2 x = π - arcsin (\frac{15}{17}) + 2 πn

解

2 x = arcsin (\frac{15}{17}) + 2 πn : x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

2 x = arcsin (\frac{15}{17}) + 2 πn

两边除以

2

2 x = arcsin (\frac{15}{17}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{15}{17} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

解

2 x = π - arcsin (\frac{15}{17}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

2 x = π - arcsin (\frac{15}{17}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{15}{17}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{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{15}{17} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn, x = \frac{3 π}{4} + πn

将解代入原方程进行验证

将它们代入

tan (2 x) + sec (2 x) = 4

检验解是否符合

去除与方程不符的解。

检验

\frac{arcsin ( \frac{15}{17} )}{2} + πn

的解:

真

\frac{arcsin ( \frac{15}{17} )}{2} + πn

代入

n = 1

\frac{arcsin ( \frac{15}{17} )}{2} + π 1

对于

tan (2 x) + sec (2 x) = 4

代入

x = \frac{arcsin ( \frac{15}{17} )}{2} + π 1

tan (2 (\frac{arcsin ( \frac{15}{17} )}{2} + π 1)) + sec (2 (\frac{arcsin ( \frac{15}{17} )}{2} + π 1)) = 4

整理后得

4 = 4

\Rightarrow 真

检验

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

的解:

假

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + πn

代入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1

对于

tan (2 x) + sec (2 x) = 4

代入

x = \frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1

tan (2 (\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1)) + sec (2 (\frac{π}{2} - \frac{arcsin ( \frac{15}{17} )}{2} + π 1)) = 4

整理后得

- 4 = 4

\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

tan (2 x) + sec (2 x) = 4

代入

x = \frac{3 π}{4} + π 1

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

未定义

\Rightarrow 假

x = \frac{arcsin ( \frac{15}{17} )}{2} + πn

以小数形式表示解

x = \frac{1.08083 \dots}{2} + πn

作图

流行的例子

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0=-2sin(x)-4cos(2x)

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- 0.1429 = cos (C)

sin (4 θ) = 0.345

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