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受欢迎的三角函数 >

1-tan(x)=sec(x)

解答

1 - tan (x) = sec (x)

解答

x = 2 πn, x = 2 π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 36 0^{\circ} + 36 0^{\circ} n

求解步骤

1 - tan (x) = sec (x)

两边减去

sec (x)

1 - tan (x) - sec (x) = 0

用 sin, cos 表示

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

化简

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

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

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

两边加上

sin (x)

cos (x) - 1 = sin (x)

两边进行平方

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

两边减去

sin^{2} (x)

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

分解

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

整理后得

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

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

分别求解每个部分

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

cos (x) - 1 + sin (x) = 0 : x = 2 πn, x = 2 πn + \frac{π}{2}

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

使用三角恒等式改写

cos (x) - 1 + sin (x)

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

sin (x) + cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

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

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= - 1 + 2 sin (x + \frac{π}{4})

- 1 + 2 sin (x + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (x + \frac{π}{4}) = 0

两边加上

1

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

化简

2 sin (x + \frac{π}{4}) = 1

2 sin (x + \frac{π}{4}) = 1

两边除以

2

2 sin (x + \frac{π}{4}) = 1

两边除以

2

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

化简

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

化简

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

约分:

2

= sin (x + \frac{π}{4})

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

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{π}{4} = \frac{π}{4} + 2 πn, x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

x + \frac{π}{4} = \frac{π}{4} + 2 πn, x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

解

x + \frac{π}{4} = \frac{π}{4} + 2 πn : x = 2 πn

x + \frac{π}{4} = \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = \frac{π}{4} + 2 πn - \frac{π}{4}

化简

x = 2 πn

解

x + \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = 2 πn + \frac{π}{2}

x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= x

化简

\frac{3 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{2}

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

x = 2 πn + \frac{π}{2}

x = 2 πn + \frac{π}{2}

x = 2 πn + \frac{π}{2}

x = 2 πn, x = 2 πn + \frac{π}{2}

cos (x) - 1 - sin (x) = 0 : x = \frac{3 π}{2} + 2 πn, x = 2 π + 2 πn

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

使用三角恒等式改写

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

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

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

使用和差化积恒等式:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

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

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

2 sin (\frac{x + \frac{π}{2} - x}{2}) sin (\frac{x - ( \frac{π}{2} - x )}{2})

\frac{x + \frac{π}{2} - x}{2} = \frac{π}{4}

\frac{x + \frac{π}{2} - x}{2}

x + \frac{π}{2} - x = \frac{π}{2}

x + \frac{π}{2} - x

对同类项分组

= x - x + \frac{π}{2}

同类项相加:

x - x = 0

= \frac{π}{2}

= \frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

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

\frac{x - ( \frac{π}{2} - x )}{2} = \frac{4 x - π}{4}

\frac{x - ( \frac{π}{2} - x )}{2}

乘开

x - (\frac{π}{2} - x) : 2 x - \frac{π}{2}

x - (\frac{π}{2} - x)

- (\frac{π}{2} - x) : - \frac{π}{2} + x

- (\frac{π}{2} - x)

打开括号

= - (\frac{π}{2}) - (- x)

使用加减运算法则

- (- a) = a, - (a) = - a

= - \frac{π}{2} + x

= x - \frac{π}{2} + x

化简

x - \frac{π}{2} + x : 2 x - \frac{π}{2}

x - \frac{π}{2} + x

对同类项分组

= x + x - \frac{π}{2}

同类项相加:

x + x = 2 x

= 2 x - \frac{π}{2}

= 2 x - \frac{π}{2}

= \frac{2 x - \frac{π}{2}}{2}

化简

2 x - \frac{π}{2} : \frac{4 x - π}{2}

2 x - \frac{π}{2}

将项转换为分式:

2 x = \frac{2 x 2}{2}

= \frac{2 x \cdot 2}{2} - \frac{π}{2}

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 x \cdot 2 - π}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 x - π}{2}

= \frac{\frac{4 x - π}{2}}{2}

使用分式法则:

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

= \frac{4 x - π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 x - π}{4}

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

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

= 2 \cdot \frac{2}{2} sin (\frac{4 x - π}{4})

分式相乘:

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

= \frac{2 \cdot 2 sin ( \frac{4 x - π}{4} )}{2}

约分:

2

= 2 sin (\frac{4 x - π}{4})

= - 1 - 2 sin (\frac{4 x - π}{4})

- 1 - 2 sin (\frac{4 x - π}{4}) = 0

将

1

到右边

- 1 - 2 sin (\frac{4 x - π}{4}) = 0

两边加上

1

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

化简

- 2 sin (\frac{4 x - π}{4}) = 1

- 2 sin (\frac{4 x - π}{4}) = 1

两边除以

- 2

- 2 sin (\frac{4 x - π}{4}) = 1

两边除以

- 2

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

化简

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

化简

\frac{- 2 sin ( \frac{4 x - π}{4} )}{- 2} : sin (\frac{4 x - π}{4})

\frac{- 2 sin ( \frac{4 x - π}{4} )}{- 2}

使用分式法则:

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

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

约分:

2

= sin (\frac{4 x - π}{4})

化简

\frac{1}{- 2} : - \frac{2}{2}

\frac{1}{- 2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

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

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

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}

\frac{4 x - π}{4} = \frac{5 π}{4} + 2 πn, \frac{4 x - π}{4} = \frac{7 π}{4} + 2 πn

\frac{4 x - π}{4} = \frac{5 π}{4} + 2 πn, \frac{4 x - π}{4} = \frac{7 π}{4} + 2 πn

解

\frac{4 x - π}{4} = \frac{5 π}{4} + 2 πn : x = \frac{3 π}{2} + 2 πn

\frac{4 x - π}{4} = \frac{5 π}{4} + 2 πn

在两边乘以

4

\frac{4 x - π}{4} = \frac{5 π}{4} + 2 πn

在两边乘以

4

\frac{4 ( 4 x - π )}{4} = 4 \cdot \frac{5 π}{4} + 4 \cdot 2 πn

化简

\frac{4 ( 4 x - π )}{4} = 4 \cdot \frac{5 π}{4} + 4 \cdot 2 πn

化简

\frac{4 ( 4 x - π )}{4} : 4 x - π

\frac{4 ( 4 x - π )}{4}

数字相除:

\frac{4}{4} = 1

= 4 x - π

化简

4 \cdot \frac{5 π}{4} + 4 \cdot 2 πn : 5 π + 8 πn

4 \cdot \frac{5 π}{4} + 4 \cdot 2 πn

4 \cdot \frac{5 π}{4} = 5 π

4 \cdot \frac{5 π}{4}

分式相乘:

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

= \frac{5 π 4}{4}

约分:

4

= 5 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 5 π + 8 πn

4 x - π = 5 π + 8 πn

4 x - π = 5 π + 8 πn

4 x - π = 5 π + 8 πn

将

π

到右边

4 x - π = 5 π + 8 πn

两边加上

π

4 x - π + π = 5 π + 8 πn + π

化简

4 x = 6 π + 8 πn

4 x = 6 π + 8 πn

两边除以

4

4 x = 6 π + 8 πn

两边除以

4

\frac{4 x}{4} = \frac{6 π}{4} + \frac{8 πn}{4}

化简

\frac{4 x}{4} = \frac{6 π}{4} + \frac{8 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{6 π}{4} + \frac{8 πn}{4} : \frac{3 π}{2} + 2 πn

\frac{6 π}{4} + \frac{8 πn}{4}

消掉

\frac{6 π}{4} : \frac{3 π}{2}

\frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= \frac{3 π}{2} + \frac{8 πn}{4}

数字相除:

\frac{8}{4} = 2

= \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

解

\frac{4 x - π}{4} = \frac{7 π}{4} + 2 πn : x = 2 π + 2 πn

\frac{4 x - π}{4} = \frac{7 π}{4} + 2 πn

在两边乘以

4

\frac{4 x - π}{4} = \frac{7 π}{4} + 2 πn

在两边乘以

4

\frac{4 ( 4 x - π )}{4} = 4 \cdot \frac{7 π}{4} + 4 \cdot 2 πn

化简

\frac{4 ( 4 x - π )}{4} = 4 \cdot \frac{7 π}{4} + 4 \cdot 2 πn

化简

\frac{4 ( 4 x - π )}{4} : 4 x - π

\frac{4 ( 4 x - π )}{4}

数字相除:

\frac{4}{4} = 1

= 4 x - π

化简

4 \cdot \frac{7 π}{4} + 4 \cdot 2 πn : 7 π + 8 πn

4 \cdot \frac{7 π}{4} + 4 \cdot 2 πn

4 \cdot \frac{7 π}{4} = 7 π

4 \cdot \frac{7 π}{4}

分式相乘:

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

= \frac{7 π 4}{4}

约分:

4

= 7 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 7 π + 8 πn

4 x - π = 7 π + 8 πn

4 x - π = 7 π + 8 πn

4 x - π = 7 π + 8 πn

将

π

到右边

4 x - π = 7 π + 8 πn

两边加上

π

4 x - π + π = 7 π + 8 πn + π

化简

4 x = 8 π + 8 πn

4 x = 8 π + 8 πn

两边除以

4

4 x = 8 π + 8 πn

两边除以

4

\frac{4 x}{4} = \frac{8 π}{4} + \frac{8 πn}{4}

化简

x = 2 π + 2 πn

x = 2 π + 2 πn

x = \frac{3 π}{2} + 2 πn, x = 2 π + 2 πn

合并所有解

x = 2 πn, x = 2 πn + \frac{π}{2}, x = \frac{3 π}{2} + 2 πn, x = 2 π + 2 πn

将解代入原方程进行验证

将它们代入

1 - tan (x) = sec (x)

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

1 - tan (x) = sec (x)

代入

x = 2 π 1

1 - tan (2 π 1) = sec (2 π 1)

整理后得

1 = 1

\Rightarrow 真

检验

2 πn + \frac{π}{2}

的解:

假

2 πn + \frac{π}{2}

代入

n = 1

2 π 1 + \frac{π}{2}

对于

1 - tan (x) = sec (x)

代入

x = 2 π 1 + \frac{π}{2}

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

整理后得

- \infty = \infty

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

真

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

1 - tan (x) = sec (x)

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

- \infty = - \infty

\Rightarrow 真

检验

2 π + 2 πn

的解:

真

2 π + 2 πn

代入

n = 1

2 π + 2 π 1

对于

1 - tan (x) = sec (x)

代入

x = 2 π + 2 π 1

1 - tan (2 π + 2 π 1) = sec (2 π + 2 π 1)

整理后得

1 = 1

\Rightarrow 真

x = 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 π + 2 πn

因为方程对以下值无定义:

\frac{3 π}{2} + 2 πn

x = 2 πn, x = 2 π + 2 πn

作图

流行的例子

sec((3θ)/2)=-2,0<= θ<2pi

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

tan(x-pi/6)=sqrt(3)

tan (x - \frac{π}{6}) = 3

2sin(x)tan(x)+5=4sec(x)

2 sin (x) tan (x) + 5 = 4 sec (x)

cos^2(x)-cos(x)-2=0,0<= x<= 2pi

cos^{2} (x) - cos (x) - 2 = 0, 0 \leq x \leq 2 π

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