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受欢迎的三角函数 >

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

解答

2 sec (2 x) + tan (2 x) - 3 = 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

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

用 sin, cos 表示

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

化简

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

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

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

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

两边加上

3 cos (2 x)

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

两边进行平方

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

两边减去

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

(2 + sin (2 x))^{2} - 9 cos^{2} (2 x) = 0

使用三角恒等式改写

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

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

化简

(2 + sin (2 x))^{2} - 9 (1 - sin^{2} (2 x)) : 10 sin^{2} (2 x) + 4 sin (2 x) - 5

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

(2 + sin (2 x))^{2} : 4 + 4 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 = 2, b = sin (2 x)

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

化简

2^{2} + 2 \cdot 2 sin (2 x) + sin^{2} (2 x) : 4 + 4 sin (2 x) + sin^{2} (2 x)

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

2^{2} = 4

= 4 + 2 \cdot 2 sin (2 x) + sin^{2} (2 x)

数字相乘:

2 \cdot 2 = 4

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

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

= 4 + 4 sin (2 x) + sin^{2} (2 x) - 9 (1 - sin^{2} (2 x))

乘开

- 9 (1 - sin^{2} (2 x)) : - 9 + 9 sin^{2} (2 x)

- 9 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (2 x)

= - 9 \cdot 1 - (- 9) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (2 x)

数字相乘:

9 \cdot 1 = 9

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

= 4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x)

化简

4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x) : 10 sin^{2} (2 x) + 4 sin (2 x) - 5

4 + 4 sin (2 x) + sin^{2} (2 x) - 9 + 9 sin^{2} (2 x)

对同类项分组

= 4 sin (2 x) + sin^{2} (2 x) + 9 sin^{2} (2 x) + 4 - 9

同类项相加:

sin^{2} (2 x) + 9 sin^{2} (2 x) = 10 sin^{2} (2 x)

= 4 sin (2 x) + 10 sin^{2} (2 x) + 4 - 9

数字相加/相减:

4 - 9 = - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

= 10 sin^{2} (2 x) + 4 sin (2 x) - 5

- 5 + 10 sin^{2} (2 x) + 4 sin (2 x) = 0

用替代法求解

- 5 + 10 sin^{2} (2 x) + 4 sin (2 x) = 0

令

: sin (2 x) = u

- 5 + 10 u^{2} + 4 u = 0

- 5 + 10 u^{2} + 4 u = 0 : u = \frac{- 2 + 3 6}{10}, u = - \frac{2 + 3 6}{10}

- 5 + 10 u^{2} + 4 u = 0

改写成标准形式

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

10 u^{2} + 4 u - 5 = 0

使用求根公式求解

10 u^{2} + 4 u - 5 = 0

二次方程求根公式:

若

a = 10, b = 4, c = - 5

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

4^{2} - 4 \cdot 10 (- 5) = 66

4^{2} - 4 \cdot 10 (- 5)

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 10 \cdot 5

数字相乘:

4 \cdot 10 \cdot 5 = 200

= 4^{2} + 200

4^{2} = 16

= 16 + 200

数字相加:

16 + 200 = 216

= 216

216

质因数分解:

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

216

216

除以

2216 = 108 \cdot 2

= 2 \cdot 108

108

除以

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 54

54

除以

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 27

27

除以

327 = 9 \cdot 3

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

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \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

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

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

使用指数法则:

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

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

使用根式运算法则:

n ab = n a n b

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

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 2 \cdot 3

整理后得

= 66

u_{1, 2} = \frac{- 4 \pm 6 6}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- 4 + 6 6}{2 \cdot 10}, u_{2} = \frac{- 4 - 6 6}{2 \cdot 10}

u = \frac{- 4 + 6 6}{2 \cdot 10} : \frac{- 2 + 3 6}{10}

\frac{- 4 + 6 6}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 4 + 6 6}{20}

分解

- 4 + 66 : 2 (- 2 + 36)

- 4 + 66

改写为

= - 2 \cdot 2 + 2 \cdot 36

因式分解出通项

2

= 2 (- 2 + 36)

= \frac{2 ( - 2 + 3 6 )}{20}

约分:

2

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

u = \frac{- 4 - 6 6}{2 \cdot 10} : - \frac{2 + 3 6}{10}

\frac{- 4 - 6 6}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 4 - 6 6}{20}

分解

- 4 - 66 : - 2 (2 + 36)

- 4 - 66

改写为

= - 2 \cdot 2 - 2 \cdot 36

因式分解出通项

2

= - 2 (2 + 36)

= - \frac{2 ( 2 + 3 6 )}{20}

约分:

2

= - \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

2 sec (2 x) + tan (2 x) - 3 = 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

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

代入

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

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

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

代入

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

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

整理后得

- 6 = 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

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

代入

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

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

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

代入

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

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

整理后得

- 6 = 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

流行的例子

solvefor x,5=9.4cos(x)

so l v e f or x, 5 = 9.4 cos (x)

20=2.891sin(0.016d-1.183)+18.512

20 = 2.891 sin (0.016 d - 1.183) + 18.512

2sin(θ)=-1,0<= θ<2pi

2 sin (θ) = - 1, 0 \leq θ < 2 π

cos (16 x) = 0

sin(θ)=-(sqrt(35))/6

sin (θ) = - \frac{35}{6}