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tan(x/2)-csc(x)=0

解答

tan (\frac{x}{2}) - csc (x) = 0

解答

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{5 π}{2} + 4 πn, x = \frac{7 π}{2} + 4 πn, x = \frac{π}{2} + 4 πn, x = \frac{3 π}{2} + 4 πn

+1

度数

x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n, x = 45 0^{\circ} + 72 0^{\circ} n, x = 63 0^{\circ} + 72 0^{\circ} n, x = 9 0^{\circ} + 72 0^{\circ} n, x = 27 0^{\circ} + 72 0^{\circ} n

求解步骤

tan (\frac{x}{2}) - csc (x) = 0

令

: u = \frac{x}{2}

tan (u) - csc (2 u) = 0

用 sin, cos 表示

- csc (2 u) + tan (u)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

sin (2 u), cos (u)

的最小公倍数:

sin (2 u) cos (u)

sin (2 u), cos (u)

最小公倍数 (LCM)

计算出由出现在

sin (2 u)

或

cos (u)

中的因子组成的表达式

= sin (2 u) cos (u)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (2 u) cos (u)

对于

\frac{1}{sin ( 2 u )} :

将分母和分子乘以

cos (u)

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

对于

\frac{sin ( u )}{cos ( u )} :

将分母和分子乘以

sin (2 u)

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

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

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

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

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

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

使用三角恒等式改写

- cos (u) + sin (2 u) sin (u)

使用倍角公式:

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

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

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

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

分解

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

- cos (u) + 2 cos (u) sin^{2} (u)

因式分解出通项

cos (u)

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

分解

2 sin^{2} (u) - 1 : (2 sin (u) + 1) (2 sin (u) - 1)

2 sin^{2} (u) - 1

将

2 sin^{2} (u) - 1

改写为

(2 sin (u))^{2} - 1^{2}

2 sin^{2} (u) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{2} (u) - 1

将

1

改写为

1^{2}

= (2)^{2} sin^{2} (u) - 1^{2}

使用指数法则:

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

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

= (2 sin (u))^{2} - 1^{2}

= (2 sin (u))^{2} - 1^{2}

使用平方差公式:

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

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

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

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

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

分别求解每个部分

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

2 sin (u) + 1 = 0 : u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

2 sin (u) + 1 = 0

将

1

到右边

2 sin (u) + 1 = 0

两边减去

1

2 sin (u) + 1 - 1 = 0 - 1

化简

2 sin (u) = - 1

2 sin (u) = - 1

两边除以

2

2 sin (u) = - 1

两边除以

2

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

化简

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

化简

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

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

约分:

2

= sin (u)

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

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

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

sin (u) = - \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}

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

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

2 sin (u) - 1 = 0 : u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

2 sin (u) - 1 = 0

将

1

到右边

2 sin (u) - 1 = 0

两边加上

1

2 sin (u) - 1 + 1 = 0 + 1

化简

2 sin (u) = 1

2 sin (u) = 1

两边除以

2

2 sin (u) = 1

两边除以

2

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

化简

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

化简

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

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

约分:

2

= sin (u)

化简

\frac{1}{2} : \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}

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

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

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

sin (u) = \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}

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

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

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn, u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 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{5 π}{4} + 2 πn : x = \frac{5 π}{2} + 4 πn

\frac{x}{2} = \frac{5 π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{5 π}{4} + 2 πn

在两边乘以

2

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

化简

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

2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{4} = \frac{5 π}{2}

2 \cdot \frac{5 π}{4}

分式相乘:

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

= \frac{5 π 2}{4}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{4}

约分:

2

= \frac{5 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{2} + 4 πn

x = \frac{5 π}{2} + 4 πn

x = \frac{5 π}{2} + 4 πn

x = \frac{5 π}{2} + 4 πn

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

\frac{x}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

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

化简

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

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{4} = \frac{7 π}{2}

2 \cdot \frac{7 π}{4}

分式相乘:

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

= \frac{7 π 2}{4}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{4}

约分:

2

= \frac{7 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

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

\frac{x}{2} = \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{4} + 2 πn

在两边乘以

2

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

化简

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

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

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

\frac{x}{2} = \frac{3 π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{3 π}{4} + 2 πn

在两边乘以

2

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

化简

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

2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{4} = \frac{3 π}{2}

2 \cdot \frac{3 π}{4}

分式相乘:

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

= \frac{3 π 2}{4}

数字相乘:

3 \cdot 2 = 6

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{5 π}{2} + 4 πn, x = \frac{7 π}{2} + 4 πn, x = \frac{π}{2} + 4 πn, x = \frac{3 π}{2} + 4 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{5 π}{2} + 4 πn, x = \frac{7 π}{2} + 4 πn, x = \frac{π}{2} + 4 πn, x = \frac{3 π}{2} + 4 πn

流行的例子

solvefor x,f=e^{sin(x^2)}求解

x, f = e^{s i n (x^{2})}

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5 cos (x) - 10 sin (x) cos (x) = 0

3 cos (x) + 5 = 7

3cos^2(x)+6cos(x)=-3

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

sin(2x+4)=cos(46)

sin (2 x + 4) = cos (4 6^{\circ})