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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2

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

化简

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

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

将项转换为分式:

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

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

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

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

sin (x) + cot (x) (1 + cos (x)) - 2 (1 + cos (x)) = 0

用 sin, cos 表示

sin (x) - (1 + cos (x)) \cdot 2 + (1 + cos (x)) cot (x)

使用基本三角恒等式:

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

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

化简

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

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

乘

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

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

分式相乘:

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

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

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

将项转换为分式:

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

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

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

sin (x) sin (x) - (1 + cos (x)) \cdot 2 sin (x) + cos (x) (1 + cos (x))

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

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

\frac{sin ^{2} ( x ) + ( 1 + cos ( x ) ) cos ( x ) - ( 1 + cos ( x ) ) \cdot 2 sin ( x )}{sin ( x )} = 0

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

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

乘开

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

cos (x) (1 + cos (x))

使用分配律:

a (b + c) = ab + a c

a = cos (x), b = 1, c = cos (x)

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

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

化简

1 \cdot cos (x) + cos (x) cos (x) : cos (x) + cos^{2} (x)

1 \cdot cos (x) + cos (x) cos (x)

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

1 \cdot cos (x)

乘以:

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

= cos (x)

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

cos (x) cos (x)

使用指数法则:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

乘开

- 2 sin (x) (1 + cos (x)) : - 2 sin (x) - 2 sin (x) cos (x)

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

使用分配律:

a (b + c) = ab + a c

a = - 2 sin (x), b = 1, c = cos (x)

= - 2 sin (x) \cdot 1 + (- 2 sin (x)) cos (x)

使用加减运算法则

+ (- a) = - a

= - 2 \cdot 1 \cdot sin (x) - 2 sin (x) cos (x)

数字相乘:

2 \cdot 1 = 2

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x) = 0

分解

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x) : (1 - 2 sin (x)) (cos (x) + 1)

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x)

因式分解出通项

cos (x)

= 1 + cos (x) (1 - 2 sin (x)) - 2 sin (x)

改写为

= (1 - 2 sin (x)) cos (x) + 1 \cdot (1 - 2 sin (x))

因式分解出通项

(1 - 2 sin (x))

= (1 - 2 sin (x)) (cos (x) + 1)

(1 - 2 sin (x)) (cos (x) + 1) = 0

分别求解每个部分

1 - 2 sin (x) = 0 or cos (x) + 1 = 0

1 - 2 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

1 - 2 sin (x) = 0

将

1

到右边

1 - 2 sin (x) = 0

两边减去

1

1 - 2 sin (x) - 1 = 0 - 1

化简

- 2 sin (x) = - 1

- 2 sin (x) = - 1

两边除以

- 2

- 2 sin (x) = - 1

两边除以

- 2

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

化简

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

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

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

将

1

到右边

cos (x) + 1 = 0

两边减去

1

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

化简

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = π + 2 πn

因为方程对以下值无定义:

π + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

4cos(2θ)+16cos(θ)+4=9cos(θ)

4 cos (2 θ) + 16 cos (θ) + 4 = 9 cos (θ)

sin(θ)=(7pi)/3

sin (θ) = \frac{7 π}{3}

tan(x)*cos^2(x)-tan(x)=0

tan (x) \cdot cos^{2} (x) - tan (x) = 0

sin(x+pi)-sin(x)+sqrt(2)=0

sin (x + π) - sin (x) + 2 = 0

3 tan (x) = - 3