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 >

(2sin(x)-1)/(3cos(x))<= 0

解答

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

解答

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

+2

间隔符号

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

十进制

2 πn \leq x \leq 0.52359 \dots + 2 πn or 1.57079 \dots + 2 πn < x \leq 2.61799 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

求解步骤

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

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

的周期:

2 π

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

的零点和无定义点

要找到零点，将不等式设置为零

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

\frac{2 sin ( x ) - 1}{3 cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

\frac{2 sin ( x ) - 1}{3 cos ( x )} = 0, 0 \leq x < 2 π

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

2 sin (x) - 1 = 0

将

1

到右边

2 sin (x) - 1 = 0

两边加上

1

2 sin (x) - 1 + 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

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{6}, x = \frac{5 π}{6}

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}

找到分母的零解

3 cos (x) = 0

两边除以

3

3 cos (x) = 0

两边除以

3

\frac{3 cos ( x )}{3} = \frac{0}{3}

化简

cos (x) = 0

cos (x) = 0

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 < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

\frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}

确定区间

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

总结如下表：

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

确定满足所需条件的区间：

\leq 0

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

合并重叠的区间

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

0 \leq x < \frac{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}{6}

or

x = \frac{π}{6}

0 \leq x \leq \frac{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 \leq \frac{π}{6}

or

\frac{π}{2} < x < \frac{5 π}{6}

0 \leq x \leq \frac{π}{6} or \frac{π}{2} < x < \frac{5 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 \leq \frac{π}{6} or \frac{π}{2} < x < \frac{5 π}{6}

or

x = \frac{5 π}{6}

0 \leq x \leq \frac{π}{6} or \frac{π}{2} < x \leq \frac{5 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 \leq \frac{π}{6} or \frac{π}{2} < x \leq \frac{5 π}{6}

or

\frac{3 π}{2} < x < 2 π

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

or

x = 2 π

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

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

使用周期

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

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

流行的例子

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

cos^2(x)> 1/4 ,0<= x<= 2pi

cos^{2} (x) > \frac{1}{4}, 0 \leq x \leq 2 π

sin((x*pi}{(\frac{1+sqrt(5))/2)^2})>0

sin \frac{x \cdot π}{( \frac{1 + 5}{2} ) ^{2}} > 0

2sin(x)-sqrt(3)<= 0

2 sin (x) - 3 \leq 0

cos(x)<=-(sqrt(2))/2

cos (x) \leq - \frac{2}{2}