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 >

(sin(x))/(1+cos(x))=3cot(x/2)

解答

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

解答

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

+1

度数

x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

两边减去

3 cot (\frac{x}{2})

\frac{sin ( x )}{1 + cos ( x )} - 3 cot (\frac{x}{2}) = 0

化简

\frac{sin ( x )}{1 + cos ( x )} - 3 cot (\frac{x}{2}) : \frac{sin ( x ) - 3 cot ( \frac{x}{2} ) ( 1 + cos ( x ) )}{1 + cos ( x )}

\frac{sin ( x )}{1 + cos ( x )} - 3 cot (\frac{x}{2})

将项转换为分式:

3 cot (\frac{x}{2}) = \frac{3 cot ( \frac{x}{2} ) ( 1 + cos ( x ) )}{1 + cos ( x )}

= \frac{sin ( x )}{1 + cos ( x )} - \frac{3 cot ( \frac{x}{2} ) ( 1 + cos ( 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{sin ( x ) - 3 cot ( \frac{x}{2} ) ( 1 + cos ( x ) )}{1 + cos ( x )}

\frac{sin ( x ) - 3 cot ( \frac{x}{2} ) ( 1 + cos ( x ) )}{1 + cos ( x )} = 0

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

sin (x) - 3 cot (\frac{x}{2}) (1 + cos (x)) = 0

令

: u = \frac{x}{2}

sin (2 u) - 3 cot (u) (1 + cos (2 u)) = 0

用 sin, cos 表示

sin (2 u) - (1 + cos (2 u)) \cdot 3 cot (u)

使用基本三角恒等式:

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

= sin (2 u) - (1 + cos (2 u)) \cdot 3 \cdot \frac{cos ( u )}{sin ( u )}

化简

sin (2 u) - (1 + cos (2 u)) \cdot 3 \cdot \frac{cos ( u )}{sin ( u )} : \frac{sin ( 2 u ) sin ( u ) - 3 cos ( u ) ( 1 + cos ( 2 u ) )}{sin ( u )}

sin (2 u) - (1 + cos (2 u)) \cdot 3 \cdot \frac{cos ( u )}{sin ( u )}

乘

(1 + cos (2 u)) \cdot 3 \cdot \frac{cos ( u )}{sin ( u )} : \frac{3 cos ( u ) ( cos ( 2 u ) + 1 )}{sin ( u )}

(1 + cos (2 u)) \cdot 3 \cdot \frac{cos ( u )}{sin ( u )}

分式相乘:

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

= \frac{cos ( u ) ( 1 + cos ( 2 u ) ) \cdot 3}{sin ( u )}

= sin (2 u) - \frac{3 cos ( u ) ( cos ( 2 u ) + 1 )}{sin ( u )}

将项转换为分式:

sin (2 u) = \frac{sin ( 2 u ) sin ( u )}{sin ( u )}

= \frac{sin ( 2 u ) sin ( u )}{sin ( u )} - \frac{cos ( u ) ( 1 + cos ( 2 u ) ) \cdot 3}{sin ( u )}

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{sin ( 2 u ) sin ( u ) - cos ( u ) ( 1 + cos ( 2 u ) ) \cdot 3}{sin ( u )}

= \frac{sin ( 2 u ) sin ( u ) - 3 cos ( u ) ( 1 + cos ( 2 u ) )}{sin ( u )}

\frac{sin ( 2 u ) sin ( u ) - ( 1 + cos ( 2 u ) ) \cdot 3 cos ( u )}{sin ( u )} = 0

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

sin (2 u) sin (u) - (1 + cos (2 u)) \cdot 3 cos (u) = 0

使用三角恒等式改写

sin (2 u) sin (u) - (1 + cos (2 u)) \cdot 3 cos (u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (u) cos (u) sin (u) - 3 cos (u) (1 + cos (2 u))

2 sin (u) cos (u) sin (u) = 2 sin^{2} (u) cos (u)

2 sin (u) cos (u) sin (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (u) sin (u) = sin^{1 + 1} (u)

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

数字相加:

1 + 1 = 2

= 2 cos (u) sin^{2} (u)

= 2 sin^{2} (u) cos (u) - 3 cos (u) (1 + cos (2 u))

- (1 + cos (2 u)) \cdot 3 cos (u) + 2 cos (u) sin^{2} (u) = 0

分解

- (1 + cos (2 u)) \cdot 3 cos (u) + 2 cos (u) sin^{2} (u) : cos (u) (- 3 (1 + cos (2 u)) + 2 sin^{2} (u))

- (1 + cos (2 u)) \cdot 3 cos (u) + 2 cos (u) sin^{2} (u)

因式分解出通项

cos (u)

= cos (u) (- 3 (1 + cos (2 u)) + 2 sin^{2} (u))

cos (u) (- 3 (1 + cos (2 u)) + 2 sin^{2} (u)) = 0

分别求解每个部分

cos (u) = 0 or - 3 (1 + cos (2 u)) + 2 sin^{2} (u) = 0

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

cos (u) = 0

cos (u) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

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

- 3 (1 + cos (2 u)) + 2 sin^{2} (u) = 0

使用三角恒等式改写

- (1 + cos (2 u)) \cdot 3 + 2 sin^{2} (u)

使用倍角公式:

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

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

= - (cos (2 u) - 1) - 3 (1 + cos (2 u))

化简

- (cos (2 u) - 1) - 3 (1 + cos (2 u)) : - 4 cos (2 u) - 2

- (cos (2 u) - 1) - 3 (1 + cos (2 u))

- (cos (2 u) - 1) : - cos (2 u) + 1

- (cos (2 u) - 1)

打开括号

= - (cos (2 u)) - (- 1)

使用加减运算法则

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

= - cos (2 u) + 1

= - cos (2 u) + 1 - 3 (1 + cos (2 u))

乘开

- 3 (1 + cos (2 u)) : - 3 - 3 cos (2 u)

- 3 (1 + cos (2 u))

使用分配律:

a (b + c) = ab + a c

a = - 3, b = 1, c = cos (2 u)

= - 3 \cdot 1 + (- 3) cos (2 u)

使用加减运算法则

+ (- a) = - a

= - 3 \cdot 1 - 3 cos (2 u)

数字相乘:

3 \cdot 1 = 3

= - 3 - 3 cos (2 u)

= - cos (2 u) + 1 - 3 - 3 cos (2 u)

化简

- cos (2 u) + 1 - 3 - 3 cos (2 u) : - 4 cos (2 u) - 2

- cos (2 u) + 1 - 3 - 3 cos (2 u)

对同类项分组

= - cos (2 u) - 3 cos (2 u) + 1 - 3

同类项相加:

- cos (2 u) - 3 cos (2 u) = - 4 cos (2 u)

= - 4 cos (2 u) + 1 - 3

数字相加/相减:

1 - 3 = - 2

= - 4 cos (2 u) - 2

= - 4 cos (2 u) - 2

= - 4 cos (2 u) - 2

- 2 - 4 cos (2 u) = 0

将

2

到右边

- 2 - 4 cos (2 u) = 0

两边加上

2

- 2 - 4 cos (2 u) + 2 = 0 + 2

化简

- 4 cos (2 u) = 2

- 4 cos (2 u) = 2

两边除以

- 4

- 4 cos (2 u) = 2

两边除以

- 4

\frac{- 4 cos ( 2 u )}{- 4} = \frac{2}{- 4}

化简

\frac{- 4 cos ( 2 u )}{- 4} = \frac{2}{- 4}

化简

\frac{- 4 cos ( 2 u )}{- 4} : cos (2 u)

\frac{- 4 cos ( 2 u )}{- 4}

使用分式法则:

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

= \frac{4 cos ( 2 u )}{4}

数字相除:

\frac{4}{4} = 1

= cos (2 u)

化简

\frac{2}{- 4} : - \frac{1}{2}

\frac{2}{- 4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

cos (2 u) = - \frac{1}{2}

cos (2 u) = - \frac{1}{2}

cos (2 u) = - \frac{1}{2}

cos (2 u) = - \frac{1}{2}

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

解

2 u = \frac{2 π}{3} + 2 πn : u = \frac{π}{3} + πn

2 u = \frac{2 π}{3} + 2 πn

两边除以

2

2 u = \frac{2 π}{3} + 2 πn

两边除以

2

\frac{2 u}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 u}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

数字相除:

\frac{2}{2} = 1

= u

化简

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

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

使用分式法则:

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

= \frac{2 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π}{6}

约分:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

u = \frac{π}{3} + πn

u = \frac{π}{3} + πn

u = \frac{π}{3} + πn

解

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

2 u = \frac{4 π}{3} + 2 πn

两边除以

2

2 u = \frac{4 π}{3} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

数字相除:

\frac{2}{2} = 1

= u

化简

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

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

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

使用分式法则:

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

= \frac{4 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{4 π}{6}

约分:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

u = \frac{2 π}{3} + πn

u = \frac{2 π}{3} + πn

u = \frac{2 π}{3} + πn

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

合并所有解

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

u = \frac{x}{2}

代回

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

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

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

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

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

\frac{x}{2} = \frac{π}{3} + πn

在两边乘以

2

\frac{x}{2} = \frac{π}{3} + πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{3} + 2 πn : \frac{2 π}{3} + 2 πn

2 \cdot \frac{π}{3} + 2 πn

乘

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

分式相乘:

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

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn

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

\frac{x}{2} = \frac{2 π}{3} + πn

在两边乘以

2

\frac{x}{2} = \frac{2 π}{3} + πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{2 π}{3} + 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

分式相乘:

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

= \frac{2 π 2}{3}

数字相乘:

2 \cdot 2 = 4

= \frac{4 π}{3}

= \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

x = \frac{4 π}{3} + 2 πn

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

因为方程对以下值无定义:

π + 4 πn, 3 π + 4 πn

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

作图

流行的例子

sin(2x)=(sqrt(3))/2 ,0<= x<= 360

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

3sin^2(x)+8cos^2(x)=6

3 sin^{2} (x) + 8 cos^{2} (x) = 6

tan (θ) = \frac{2}{8}

4tan^2(x)+7tan(x)-4=0,cot(2x)

4 tan^{2} (x) + 7 tan (x) - 4 = 0, cot (2 x)

11.75=5sin((pit)/6-(7pi)/6)+8

11.75 = 5 sin (\frac{π t}{6} - \frac{7 π}{6}) + 8