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(x)-cot(x)>0

解答

tan (x) - cot (x) > 0

解答

\frac{π}{4} + πn < x < \frac{π}{2} + πn or \frac{3 π}{4} + πn < x < π + πn

+2

间隔符号

(\frac{π}{4} + πn, \frac{π}{2} + πn) \cup (\frac{3 π}{4} + πn, π + πn)

十进制

0.78539 \dots + πn < x < 1.57079 \dots + πn or 2.35619 \dots + πn < x < 3.14159 \dots + πn

求解步骤

tan (x) - cot (x) > 0

tan (x) - cot (x)

的周期:

π

周期函数和的复合周期是这些周期的最小公倍数

tan (x), cot (x)

tan (x)

的周期:

π

tan (x)

的周期是

π

= π

cot (x)

的周期:

π

cot (x)

的周期是

π

= π

合并周期:

π, π

= π

用 sin, cos 表示

tan (x) - cot (x) > 0

使用基本三角恒等式:

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

\frac{sin ( x )}{cos ( x )} - cot (x) > 0

使用基本三角恒等式:

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

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} > 0

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} > 0

化简

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

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )}

cos (x), sin (x)

的最小公倍数:

cos (x) sin (x)

cos (x), sin (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

sin (x)

中的因子组成的表达式

= cos (x) sin (x)

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

cos (x) sin (x)

对于

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

将分母和分子乘以

sin (x)

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

对于

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

将分母和分子乘以

cos (x)

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

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

\frac{sin ^{2} ( x ) - cos ^{2} ( x )}{cos ( x ) sin ( x )} > 0

确定

0 \leq x < π

时

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

的零点和无定义点

要找到零点，将不等式设置为零

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

\frac{sin ^{2} ( x ) - cos ^{2} ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < π : x = \frac{π}{4}, x = \frac{3 π}{4}

\frac{sin ^{2} ( x ) - cos ^{2} ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < π

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

= - cos (2 x)

- cos (2 x) = 0

两边除以

- 1

- cos (2 x) = 0

两边除以

- 1

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

化简

cos (2 x) = 0

cos (2 x) = 0

cos (2 x) = 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}

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

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

解

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

2 x = \frac{π}{2} + 2 πn

两边除以

2

2 x = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

解

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

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

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

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

在

0 \leq x < π

范围内的解

x = \frac{π}{4}, x = \frac{3 π}{4}

确定无定义点:

x = \frac{π}{2}, x = 0

找到分母的零解

cos (x) sin (x) = 0

分别求解每个部分

cos (x) = 0 or sin (x) = 0

cos (x) = 0, 0 \leq x < π : x = \frac{π}{2}

cos (x) = 0, 0 \leq x < π

cos (x) = 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}

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

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

在

0 \leq x < π

范围内的解

x = \frac{π}{2}

sin (x) = 0, 0 \leq x < π : x = 0

sin (x) = 0, 0 \leq x < π

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

在

0 \leq x < π

范围内的解

x = 0

合并所有解

x = \frac{π}{2}, x = 0

0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}

确定区间

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < π

总结如下表：

sin^{2} (x) - cos^{2} (x) cos (x) sin (x) \frac{s i n ^{2} ( x ) - c o s ^{2} ( x )}{c o s ( x ) s i n ( x )} x = 0 - + 0 未定义 0 < x < \frac{π}{4} - + + - x = \frac{π}{4} 0 + + 0 \frac{π}{4} < x < \frac{π}{2} + + + + x = \frac{π}{2} + 0 + 未定义 \frac{π}{2} < x < \frac{3 π}{4} + - + - x = \frac{3 π}{4} 0 - + 0 \frac{3 π}{4} < x < π - - + + x = π - - 0 未定义

确定满足所需条件的区间：

> 0

\frac{π}{4} < x < \frac{π}{2} or \frac{3 π}{4} < x < π

使用周期

tan (x) - cot (x)

\frac{π}{4} + πn < x < \frac{π}{2} + πn or \frac{3 π}{4} + πn < x < π + πn

流行的例子

sin^2(x)+cos(x)>= 1

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

cos(x)>= (sqrt(3))/2

cos (x) \geq \frac{3}{2}

cos (x) - 1 \geq 0

2 sin (\frac{x}{2}) - 1 > 0

sin(3x)<(sqrt(2))/2

sin (3 x) < \frac{2}{2}