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 >

5sec(x)+5tan(x)=5

解答

5 sec (x) + 5 tan (x) = 5

解答

x = 2 πn + 2 π

+1

度数

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

求解步骤

5 sec (x) + 5 tan (x) = 5

两边减去

5

5 sec (x) + 5 tan (x) - 5 = 0

用 sin, cos 表示

5 \cdot \frac{1}{cos ( x )} + 5 \cdot \frac{sin ( x )}{cos ( x )} - 5 = 0

化简

5 \cdot \frac{1}{cos ( x )} + 5 \cdot \frac{sin ( x )}{cos ( x )} - 5 : \frac{5 + 5 sin ( x ) - 5 cos ( x )}{cos ( x )}

5 \cdot \frac{1}{cos ( x )} + 5 \cdot \frac{sin ( x )}{cos ( x )} - 5

5 \cdot \frac{1}{cos ( x )} = \frac{5}{cos ( x )}

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

分式相乘:

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

= \frac{1 \cdot 5}{cos ( x )}

数字相乘:

1 \cdot 5 = 5

= \frac{5}{cos ( x )}

5 \cdot \frac{sin ( x )}{cos ( x )} = \frac{5 sin ( x )}{cos ( x )}

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

分式相乘:

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

= \frac{sin ( x ) \cdot 5}{cos ( x )}

= \frac{5}{cos ( x )} + \frac{5 sin ( x )}{cos ( x )} - 5

合并分式

\frac{5}{cos ( x )} + \frac{5 sin ( x )}{cos ( x )} : \frac{5 + 5 sin ( x )}{cos ( x )}

使用法则

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

= \frac{5 + 5 sin ( x )}{cos ( x )}

= \frac{5 sin ( x ) + 5}{cos ( x )} - 5

将项转换为分式:

5 = \frac{5 cos ( x )}{cos ( x )}

= \frac{5 + sin ( x ) \cdot 5}{cos ( x )} - \frac{5 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{5 + sin ( x ) \cdot 5 - 5 cos ( x )}{cos ( x )}

\frac{5 + 5 sin ( x ) - 5 cos ( x )}{cos ( x )} = 0

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

5 + 5 sin (x) - 5 cos (x) = 0

两边加上

5 cos (x)

5 + 5 sin (x) = 5 cos (x)

两边进行平方

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

两边减去

(5 cos (x))^{2}

(5 + 5 sin (x))^{2} - 25 cos^{2} (x) = 0

分解

(5 + 5 sin (x))^{2} - 25 cos^{2} (x) : 25 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x))

(5 + 5 sin (x))^{2} - 25 cos^{2} (x)

将

(5 + 5 sin (x))^{2} - 25 cos^{2} (x)

改写为

(5 + 5 sin (x))^{2} - (5 cos (x))^{2}

(5 + 5 sin (x))^{2} - 25 cos^{2} (x)

将

25

改写为

5^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

5^{2} cos^{2} (x) = (5 cos (x))^{2}

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

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

使用平方差公式:

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

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

= ((5 + 5 sin (x)) + 5 cos (x)) ((5 + 5 sin (x)) - 5 cos (x))

整理后得

= (5 sin (x) + 5 cos (x) + 5) (5 sin (x) - 5 cos (x) + 5)

分解

5 + 5 sin (x) + 5 cos (x) : 5 (1 + sin (x) + cos (x))

5 + 5 sin (x) + 5 cos (x)

因式分解出通项

5

= 5 (1 + sin (x) + cos (x))

= 5 (sin (x) + cos (x) + 1) (5 sin (x) - 5 cos (x) + 5)

分解

5 + 5 sin (x) - 5 cos (x) : 5 (1 + sin (x) - cos (x))

5 + 5 sin (x) - 5 cos (x)

因式分解出通项

5

= 5 (1 + sin (x) - cos (x))

= 5 (1 + sin (x) + cos (x)) \cdot 5 (1 + sin (x) - cos (x))

整理后得

= 25 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x))

25 (1 + sin (x) + cos (x)) (1 + sin (x) - cos (x)) = 0

分别求解每个部分

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

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

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

使用三角恒等式改写

1 + sin (x) + cos (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 + π

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

两边减去

\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 + π

\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} : π

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 = 4 π

= \frac{4 π}{4}

数字相除:

\frac{4}{4} = 1

= π

= 2 πn + π

x = 2 πn + π

x = 2 πn + π

x = 2 πn + π

解

x + \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = 2 πn + \frac{3 π}{2}

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

两边减去

\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 + \frac{3 π}{2}

\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} : \frac{3 π}{2}

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 = 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}

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

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

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

使用三角恒等式改写

1 + sin (x) - cos (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

合并所有解

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

将解代入原方程进行验证

将它们代入

5 sec (x) + 5 tan (x) = 5

检验解是否符合

去除与方程不符的解。

检验

2 πn + π

的解:

假

2 πn + π

代入

n = 1

2 π 1 + π

对于

5 sec (x) + 5 tan (x) = 5

代入

x = 2 π 1 + π

5 sec (2 π 1 + π) + 5 tan (2 π 1 + π) = 5

整理后得

- 5 = 5

\Rightarrow 假

检验

2 πn + \frac{3 π}{2}

的解:

假

2 πn + \frac{3 π}{2}

代入

n = 1

2 π 1 + \frac{3 π}{2}

对于

5 sec (x) + 5 tan (x) = 5

代入

x = 2 π 1 + \frac{3 π}{2}

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

未定义

\Rightarrow 假

检验

2 πn + 2 π

的解:

真

2 πn + 2 π

代入

n = 1

2 π 1 + 2 π

对于

5 sec (x) + 5 tan (x) = 5

代入

x = 2 π 1 + 2 π

5 sec (2 π 1 + 2 π) + 5 tan (2 π 1 + 2 π) = 5

整理后得

5 = 5

\Rightarrow 真

x = 2 πn + 2 π

作图

流行的例子

cos(2θ)+8sin^2(θ)=4

2cos^2(θ)-5cos(θ)+2=0

tan^2(x)-3tan(x)-4=0