zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tan(x)+sec(x)=1

解答

tan (x) + sec (x) = 1

解答

x = 2 πn + 2 π

+1

度数

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

求解步骤

tan (x) + sec (x) = 1

两边减去

1

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

用 sin, cos 表示

- 1 + sec (x) + tan (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

合并分式

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

使用法则

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

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

乘以:

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

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

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

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

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

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

使用三角恒等式改写

1 - cos (x) + 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

两边减去

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}

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

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

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

\frac{5 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{5 π}{4}

合并分式

\frac{π}{4} + \frac{5 π}{4} : \frac{3 π}{2}

使用法则

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

= \frac{π + 5 π}{4}

同类项相加:

π + 5 π = 6 π

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

x = 2 πn + \frac{3 π}{2}

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

\frac{7 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{7 π}{4}

合并分式

\frac{π}{4} + \frac{7 π}{4} : 2 π

使用法则

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

= \frac{π + 7 π}{4}

同类项相加:

π + 7 π = 8 π

= \frac{8 π}{4}

数字相除:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

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

因为方程对以下值无定义:

2 πn + \frac{3 π}{2}

x = 2 πn + 2 π

作图

流行的例子

arctan (\frac{3}{4})

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

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

tan (10 5^{\circ})

cos(2θ)=(sqrt(3))/2

cos (2 θ) = \frac{3}{2}

证明 cos(2x)=cos^2(x)-sin^2(x)

p ro v e cos (2 x) = cos^{2} (x) - sin^{2} (x)