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1/(cos(x))+tan(x)=2

解答

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

解答

x = 0.64350 \dots + 2 πn

+1

度数

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2

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

化简

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

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

将项转换为分式:

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

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

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

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

1 + tan (x) cos (x) - 2 cos (x) = 0

用 sin, cos 表示

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

化简

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

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

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

\frac{sin ( x )}{cos ( x )} cos (x)

分式相乘:

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

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

约分:

cos (x)

= sin (x)

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

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

两边加上

2 cos (x)

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

两边进行平方

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

两边减去

(2 cos (x))^{2}

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

使用三角恒等式改写

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

化简

(1 + sin (x))^{2} - 4 (1 - sin^{2} (x)) : 5 sin^{2} (x) + 2 sin (x) - 3

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

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

使用完全平方公式:

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

a = 1, b = sin (x)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

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

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

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

乘开

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

- 4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

使用加减运算法则

- (- a) = a

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

数字相乘:

4 \cdot 1 = 4

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

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

化简

1 + 2 sin (x) + sin^{2} (x) - 4 + 4 sin^{2} (x) : 5 sin^{2} (x) + 2 sin (x) - 3

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

对同类项分组

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

同类项相加:

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

= 2 sin (x) + 5 sin^{2} (x) + 1 - 4

数字相加/相减:

1 - 4 = - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

= 5 sin^{2} (x) + 2 sin (x) - 3

- 3 + 2 sin (x) + 5 sin^{2} (x) = 0

用替代法求解

- 3 + 2 sin (x) + 5 sin^{2} (x) = 0

令

: sin (x) = u

- 3 + 2 u + 5 u^{2} = 0

- 3 + 2 u + 5 u^{2} = 0 : u = \frac{3}{5}, u = - 1

- 3 + 2 u + 5 u^{2} = 0

改写成标准形式

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

5 u^{2} + 2 u - 3 = 0

使用求根公式求解

5 u^{2} + 2 u - 3 = 0

二次方程求根公式:

若

a = 5, b = 2, c = - 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 3) = 8

2^{2} - 4 \cdot 5 (- 3)

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

数字相加:

4 + 60 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- 2 + 8}{2 \cdot 5}, u_{2} = \frac{- 2 - 8}{2 \cdot 5}

u = \frac{- 2 + 8}{2 \cdot 5} : \frac{3}{5}

\frac{- 2 + 8}{2 \cdot 5}

数字相加/相减:

- 2 + 8 = 6

= \frac{6}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{6}{10}

约分:

2

= \frac{3}{5}

u = \frac{- 2 - 8}{2 \cdot 5} : - 1

\frac{- 2 - 8}{2 \cdot 5}

数字相减:

- 2 - 8 = - 10

= \frac{- 10}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 10}{10}

使用分式法则:

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

= - \frac{10}{10}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{3}{5}, u = - 1

u = sin (x)

代回

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

使用反三角函数性质

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

的通解

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

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

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

sin (x) = - 1

sin (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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{3}{5}) + 2 πn

的解:

真

arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

arcsin (\frac{3}{5}) + 2 π 1

对于

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

代入

x = arcsin (\frac{3}{5}) + 2 π 1

\frac{1}{cos ( arcsin ( \frac{3}{5} ) + 2 π 1 )} + tan (arcsin (\frac{3}{5}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

π - arcsin (\frac{3}{5}) + 2 πn

的解:

假

π - arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

对于

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

代入

x = π - arcsin (\frac{3}{5}) + 2 π 1

\frac{1}{cos ( π - arcsin ( \frac{3}{5} ) + 2 π 1 )} + tan (π - arcsin (\frac{3}{5}) + 2 π 1) = 2

整理后得

- 2 = 2

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

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

未定义

\Rightarrow 假

x = arcsin (\frac{3}{5}) + 2 πn

以小数形式表示解

x = 0.64350 \dots + 2 πn

流行的例子

-5=-5+4sin(-2x+(3pi)/4)

- 5 = - 5 + 4 sin (- 2 x + \frac{3 π}{4})

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

6 sec^{2} (x) + tan (x) - 9 = 0

3tan^2(x)-cos^2(x)=sin^2(x)

3 tan^{2} (x) - cos^{2} (x) = sin^{2} (x)

cos(θ)=(2pi)/3

cos (θ) = \frac{2 π}{3}

3(tan(θ)-2)=2tan(-7)

3 (tan (θ) - 2) = 2 tan (- 7)