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)+sqrt(2))/(cos(x))<= 0

解答

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

解答

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

+2

间隔符号

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

十进制

1.57079 \dots + 2 πn < x \leq 3.92699 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 5.49778 \dots + 2 πn

求解步骤

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

\frac{2 sin ( x ) + 2}{cos ( x )}

的周期:

2 π

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

\frac{2 sin ( x ) + 2}{cos ( x )} = 0, 0 \leq x < 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

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

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

2 sin (x) + 2 = 0

将

2

到右边

2 sin (x) + 2 = 0

两边减去

2

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

化简

2 sin (x) = - 2

2 sin (x) = - 2

两边除以

2

2 sin (x) = - 2

两边除以

2

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

化简

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

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

sin (x) = - \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}

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

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

在

0 \leq x < 2 π

范围内的解

x = \frac{5 π}{4}, x = \frac{7 π}{4}

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}

找到分母的零解

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}{2}, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}

确定区间

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

总结如下表：

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

确定满足所需条件的区间：

\leq 0

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

合并重叠的区间

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

\frac{π}{2} < x < \frac{5 π}{4}

or

x = \frac{5 π}{4}

\frac{π}{2} < x \leq \frac{5 π}{4}

两个区间的并集是指存在于任一区间的数的集合

\frac{π}{2} < x \leq \frac{5 π}{4}

or

\frac{3 π}{2} < x < \frac{7 π}{4}

\frac{π}{2} < x \leq \frac{5 π}{4} or \frac{3 π}{2} < 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

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

or

x = \frac{7 π}{4}

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

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

使用周期

\frac{2 sin ( x ) + 2}{cos ( x )}

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

流行的例子

\frac{1}{2} > cos (x)

2cos^2(a)-1>=-1/8

2 cos^{2} (a) - 1 \geq - \frac{1}{8}

4sin^2(x)-3<0,-2pi<= x<= 0

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

sin^{2} (x) > 0.5

tan (x) < 7