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受欢迎的三角函数 >

(2cos(x)-sqrt(2))/(sin(2x))<= 0

解答

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

解答

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

+2

间隔符号

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

十进制

0.78539 \dots + 2 πn \leq x < 1.57079 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 5.49778 \dots + 2 πn \leq x < 6.28318 \dots + 2 πn

求解步骤

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

利用以下特性:

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

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

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

的周期:

2 π

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

包含以下函数及对应周期:

cos (x)

的周期为

2 π

复合周期为:

= 2 π

确定

0 \leq x < 2 π

时

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

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

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

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

- 2 + 2 cos (x) = 0

将

2

到右边

- 2 + 2 cos (x) = 0

两边加上

2

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

化简

2 cos (x) = 2

2 cos (x) = 2

两边除以

2

2 cos (x) = 2

两边除以

2

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

化简

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

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

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

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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{4}, x = \frac{7 π}{4}

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

找到分母的零解

2 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 < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 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 < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 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 < 2 π

范围内的解

x = 0, x = π

合并所有解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

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

确定区间

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

确定满足所需条件的区间：

\leq 0

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

合并重叠的区间

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

or

\frac{π}{4} < x < \frac{π}{2}

\frac{π}{4} \leq x < \frac{π}{2}

两个区间的并集是指存在于任一区间的数的集合

\frac{π}{4} \leq x < \frac{π}{2}

or

π < x < \frac{3 π}{2}

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

两个区间的并集是指存在于任一区间的数的集合

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

or

x = \frac{7 π}{4}

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

or

\frac{7 π}{4} < x < 2 π

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

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

使用周期

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

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

流行的例子

((4*cos^2(x)-3))/((sin(x)+cos(x)+5))<0

\frac{( 4 \cdot cos ^{2} ( x ) - 3 )}{( sin ( x ) + cos ( x ) + 5 )} < 0

1/2 <= sin(x),cos(x)<= (sqrt(2))/2

\frac{1}{2} \leq sin (x), cos (x) \leq \frac{2}{2}

sin (θ) \leq \frac{π}{6}

tan(x)>= sin(x)

tan (x) \geq sin (x)

1 \leq tan (x)