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(θ)+sec(θ)=1

解答

tan (θ) + sec (θ) = 1

解答

θ = 2 πn + 2 π

+1

度数

θ = 36 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (θ) + sec (θ) = 1

两边减去

1

tan (θ) + sec (θ) - 1 = 0

用 sin, cos 表示

- 1 + sec (θ) + tan (θ)

使用基本三角恒等式:

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

= - 1 + \frac{1}{cos ( θ )} + tan (θ)

使用基本三角恒等式:

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

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

化简

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

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

合并分式

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

使用法则

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

= \frac{1 + sin ( θ )}{cos ( θ )}

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

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

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 ( θ ) + 1 + sin ( θ )}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

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

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

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

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

1 - cos (θ) + sin (θ) = 0

使用三角恒等式改写

1 - cos (θ) + sin (θ)

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

sin (θ) - cos (θ)

改写为

= 2 (\frac{1}{2} sin (θ) - \frac{1}{2} cos (θ))

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 (θ) - sin (\frac{π}{4}) cos (θ))

使用角和恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (θ - \frac{π}{4})

= 1 + 2 sin (θ - \frac{π}{4})

1 + 2 sin (θ - \frac{π}{4}) = 0

将

1

到右边

1 + 2 sin (θ - \frac{π}{4}) = 0

两边减去

1

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

化简

2 sin (θ - \frac{π}{4}) = - 1

2 sin (θ - \frac{π}{4}) = - 1

两边除以

2

2 sin (θ - \frac{π}{4}) = - 1

两边除以

2

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

化简

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

化简

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

\frac{2 sin ( θ - \frac{π}{4} )}{2}

约分:

2

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

sin (θ - \frac{π}{4}) = - \frac{2}{2}

sin (θ - \frac{π}{4}) = - \frac{2}{2}

sin (θ - \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}

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

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

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

化简

θ - \frac{π}{4} + \frac{π}{4} : θ

θ - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= θ

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}

θ = 2 πn + \frac{3 π}{2}

θ = 2 πn + \frac{3 π}{2}

θ = 2 πn + \frac{3 π}{2}

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

化简

θ - \frac{π}{4} + \frac{π}{4} : θ

θ - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= θ

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

θ = 2 πn + 2 π

θ = 2 πn + 2 π

θ = 2 πn + 2 π

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

因为方程对以下值无定义:

2 πn + \frac{3 π}{2}

θ = 2 πn + 2 π

作图

流行的例子

6sin(x)+sqrt(8)=0

6 sin (x) + 8 = 0

2cos(x)-7=4sec(x)

2 cos (x) - 7 = 4 sec (x)

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

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

tan^2(x)-(sqrt(3)+1)tan(x)+sqrt(3)=0

tan^{2} (x) - (3 + 1) tan (x) + 3 = 0

6 cos (θ) = 3