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 >

(1+2sin(x))/((cos(2x)))>0

解答

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

解答

2 πn \leq x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{7 π}{6} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{11 π}{6} + 2 πn < x \leq 2 π + 2 πn

+2

间隔符号

[2 πn, \frac{π}{4} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{7 π}{6} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{7 π}{4} + 2 πn) \cup (\frac{11 π}{6} + 2 πn, 2 π + 2 πn]

十进制

2 πn \leq x < 0.78539 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.66519 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 5.49778 \dots + 2 πn or 5.75958 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

求解步骤

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

利用以下特性:

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

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

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

的周期:

2 π

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

包含以下函数及对应周期:

sin (x)

的周期为

2 π

复合周期为:

= 2 π

确定

0 \leq x < 2 π

时

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

的零点和无定义点

要找到零点，将不等式设置为零

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

\frac{1 + 2 sin ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

\frac{1 + 2 sin ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0, 0 \leq x < 2 π

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{7 π}{6}, x = \frac{11 π}{6}

确定无定义点:

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

找到分母的零解

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

使用三角恒等式改写

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

使用倍角公式:

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

= cos (2 x)

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 < 2 π

范围内的解

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

\frac{π}{4}, \frac{3 π}{4}, \frac{7 π}{6}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{11 π}{6}

确定区间

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < \frac{7 π}{6}, \frac{7 π}{6} < x < \frac{5 π}{4}, \frac{5 π}{4} < x < \frac{7 π}{4}, \frac{7 π}{4} < x < \frac{11 π}{6}, \frac{11 π}{6} < x < 2 π

总结如下表：

1 + 2 sin (x) cos^{2} (x) - sin^{2} (x) \frac{1 + 2 s i n ( x )}{c o s ^{2} ( x ) - s i n ^{2} ( x )} x = 0 + + + 0 < x < \frac{π}{4} + + + x = \frac{π}{4} + 0 未定义 \frac{π}{4} < x < \frac{3 π}{4} + - - x = \frac{3 π}{4} + 0 未定义 \frac{3 π}{4} < x < \frac{7 π}{6} + + + x = \frac{7 π}{6} 0 + 0 \frac{7 π}{6} < x < \frac{5 π}{4} - + - x = \frac{5 π}{4} - 0 未定义 \frac{5 π}{4} < x < \frac{7 π}{4} - - + x = \frac{7 π}{4} - 0 未定义 \frac{7 π}{4} < x < \frac{11 π}{6} - + - x = \frac{11 π}{6} 0 + 0 \frac{11 π}{6} < x < 2 π + + + x = 2 π + + +

确定满足所需条件的区间：

> 0

x = 0 or 0 < x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π or x = 2 π

合并重叠的区间

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π or x = 2 π

两个区间的并集是指存在于任一区间的数的集合

x = 0

or

0 < x < \frac{π}{4}

0 \leq x < \frac{π}{4}

两个区间的并集是指存在于任一区间的数的集合

0 \leq x < \frac{π}{4}

or

\frac{3 π}{4} < x < \frac{7 π}{6}

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6}

两个区间的并集是指存在于任一区间的数的集合

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6}

or

\frac{5 π}{4} < x < \frac{7 π}{4}

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4}

两个区间的并集是指存在于任一区间的数的集合

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4}

or

\frac{11 π}{6} < x < 2 π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π

两个区间的并集是指存在于任一区间的数的集合

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x < 2 π

or

x = 2 π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x \leq 2 π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < \frac{7 π}{6} or \frac{5 π}{4} < x < \frac{7 π}{4} or \frac{11 π}{6} < x \leq 2 π

使用周期

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

2 πn \leq x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{7 π}{6} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn or \frac{11 π}{6} + 2 πn < x \leq 2 π + 2 πn

流行的例子

(cos^2(x)-1/2)/(tan(x)-sqrt(3))<0

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

tan(θ)>0,cot(θ)>0

tan (θ) > 0, cot (θ) > 0

6 sin (θ) \geq 0

tan(x)>\sqrt[4]{5},-pi<= x<= pi

tan (x) > 45, - π \leq x \leq π

2sin^2(x)+3sin(x)>= 2,0<= x<= 2pi

2 sin^{2} (x) + 3 sin (x) \geq 2, 0 \leq x \leq 2 π