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受欢迎的三角函数 >

1+sec(x)>= 0

解答

1 + sec (x) \geq 0

解答

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

+2

间隔符号

[2 πn, \frac{π}{2} + 2 πn) \cup x = π + 2 πn \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn]

十进制

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

求解步骤

1 + sec (x) \geq 0

1 + sec (x)

的周期:

2 π

周期

sec (b x + c) + d = \frac{sec ( x ) 的周期}{∣ b ∣}

sec (x)

的周期是

2 π

= \frac{2 π}{∣ 1 ∣}

化简

= 2 π

用 sin, cos 表示

1 + sec (x) \geq 0

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

1 + \frac{1}{cos ( x )} \geq 0

1 + \frac{1}{cos ( x )} \geq 0

化简

1 + \frac{1}{cos ( x )} : \frac{cos ( x ) + 1}{cos ( x )}

1 + \frac{1}{cos ( x )}

将项转换为分式:

1 = \frac{1 cos ( x )}{cos ( x )}

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{1}{cos ( 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{1 \cdot cos ( x ) + 1}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

= \frac{cos ( x ) + 1}{cos ( x )}

\frac{cos ( x ) + 1}{cos ( x )} \geq 0

确定

0 \leq x < 2 π

时

\frac{cos ( x ) + 1}{cos ( x )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{cos ( x ) + 1}{cos ( x )} = 0

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

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

用替代法求解

\frac{cos ( x ) + 1}{cos ( x )} = 0

令

: cos (x) = u

\frac{u + 1}{u} = 0

\frac{u + 1}{u} = 0 : u = - 1

\frac{u + 1}{u} = 0

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

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{u + 1}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = - 1

u = cos (x)

代回

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

在

0 \leq x < 2 π

范围内的解

x = π

合并所有解

x = π

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}, π, \frac{3 π}{2}

确定区间

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

总结如下表：

cos (x) + 1 cos (x) \frac{c o s ( x ) + 1}{c o s ( x )} x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 未定义 \frac{π}{2} < x < π + - - x = π 0 - 0 π < 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：

\geq 0

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

合并重叠的区间

0 \leq x < \frac{π}{2} or x = π 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}{2}

0 \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

0 \leq x < \frac{π}{2}

or

x = π

0 \leq x < \frac{π}{2} or x = 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}{2} or x = π

or

\frac{3 π}{2} < x < 2 π

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

or

x = 2 π

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

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

使用周期

1 + sec (x)

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

流行的例子

- sin (x) \leq 0.5

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

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

cos^2(x)<(sqrt(2))/2+sin^2(x)

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

tan (x) < - 1.5

- sin (x) \leq 0.9