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受欢迎的三角函数 >

sec^2(x)<= 4

解答

sec^{2} (x) \leq 4

解答

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

+2

间隔符号

[- \frac{π}{3} + 2 πn, \frac{π}{3} + 2 πn] \cup [\frac{2 π}{3} + 2 πn, \frac{4 π}{3} + 2 πn]

十进制

- 1.04719 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn or 2.09439 \dots + 2 πn \leq x \leq 4.18879 \dots + 2 πn

求解步骤

sec^{2} (x) \leq 4

用 sin, cos 表示

sec^{2} (x) \leq 4

使用基本三角恒等式:

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

(\frac{1}{cos ( x )})^{2} \leq 4

(\frac{1}{cos ( x )})^{2} \leq 4

对于

u^{n} \leq a

，若

n

为偶数，则

- n a \leq u \leq n a

- 4 \leq \frac{1}{cos ( x )} \leq 4

若

a \leq u \leq b

，则

a \leq u and u \leq b

- 4 \leq \frac{1}{cos ( x )} and \frac{1}{cos ( x )} \leq 4

- 4 \leq \frac{1}{cos ( x )} : cos (x) \leq - \frac{1}{2} or cos (x) > 0

- 4 \leq \frac{1}{cos ( x )}

交换两边

\frac{1}{cos ( x )} \geq - 4

改写为标准形式

\frac{1}{cos ( x )} \geq - 4

两边加上

4

\frac{1}{cos ( x )} + 4 \geq - 4 + 4

化简

\frac{1}{cos ( x )} + 4 \geq - 4 + 4

化简

\frac{1}{cos ( x )} + 4 : \frac{1}{cos ( x )} + 2

\frac{1}{cos ( x )} + 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{cos ( x )} + 2

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

化简

\frac{1}{cos ( x )} + 2 : \frac{1 + 2 cos ( x )}{cos ( x )}

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

将项转换为分式:

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

= \frac{1}{cos ( x )} + \frac{2 cos ( x )}{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 + 2 cos ( x )}{cos ( x )}

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

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

确定区间

确定

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

符号

确定

1 + 2 cos (x)

符号

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

1 + 2 cos (x) = 0

将

1

到右边

1 + 2 cos (x) = 0

两边减去

1

1 + 2 cos (x) - 1 = 0 - 1

化简

2 cos (x) = - 1

2 cos (x) = - 1

两边除以

2

2 cos (x) = - 1

两边除以

2

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

化简

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

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

1 + 2 cos (x) < 0 : cos (x) < - \frac{1}{2}

1 + 2 cos (x) < 0

将

1

到右边

1 + 2 cos (x) < 0

两边减去

1

1 + 2 cos (x) - 1 < 0 - 1

化简

2 cos (x) < - 1

2 cos (x) < - 1

两边除以

2

2 cos (x) < - 1

两边除以

2

\frac{2 cos ( x )}{2} < \frac{- 1}{2}

化简

cos (x) < - \frac{1}{2}

cos (x) < - \frac{1}{2}

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

1 + 2 cos (x) > 0

将

1

到右边

1 + 2 cos (x) > 0

两边减去

1

1 + 2 cos (x) - 1 > 0 - 1

化简

2 cos (x) > - 1

2 cos (x) > - 1

两边除以

2

2 cos (x) > - 1

两边除以

2

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

化简

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

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

确定

cos (x)

符号

cos (x) = 0

cos (x) < 0

cos (x) > 0

找到奇点

找到分母的零解

cos (x) : cos (x) = 0

总结如下表：

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

确定满足所需条件的区间：

\geq 0

cos (x) < - \frac{1}{2} or cos (x) = - \frac{1}{2} or cos (x) > 0

合并重叠的区间

cos (x) \leq - \frac{1}{2} or cos (x) > 0

两个区间的并集是指存在于任一区间的数的集合

cos (x) < - \frac{1}{2}

or

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

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

两个区间的并集是指存在于任一区间的数的集合

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

or

cos (x) > 0

cos (x) \leq - \frac{1}{2} or cos (x) > 0

cos (x) \leq - \frac{1}{2} or cos (x) > 0

cos (x) \leq - \frac{1}{2} or cos (x) > 0

\frac{1}{cos ( x )} \leq 4 : cos (x) < 0 or cos (x) \geq \frac{1}{2}

\frac{1}{cos ( x )} \leq 4

改写为标准形式

\frac{1}{cos ( x )} \leq 4

两边减去

4

\frac{1}{cos ( x )} - 4 \leq 4 - 4

化简

\frac{1}{cos ( x )} - 4 \leq 4 - 4

化简

\frac{1}{cos ( x )} - 4 : \frac{1}{cos ( x )} - 2

\frac{1}{cos ( x )} - 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

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

化简

\frac{1}{cos ( x )} - 2 : \frac{1 - 2 cos ( x )}{cos ( x )}

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

将项转换为分式:

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

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

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

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

确定区间

确定

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

符号

确定

1 - 2 cos (x)

符号

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

1 - 2 cos (x) = 0

将

1

到右边

1 - 2 cos (x) = 0

两边减去

1

1 - 2 cos (x) - 1 = 0 - 1

化简

- 2 cos (x) = - 1

- 2 cos (x) = - 1

两边除以

- 2

- 2 cos (x) = - 1

两边除以

- 2

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

化简

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

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

1 - 2 cos (x) < 0 : cos (x) > \frac{1}{2}

1 - 2 cos (x) < 0

将

1

到右边

1 - 2 cos (x) < 0

两边减去

1

1 - 2 cos (x) - 1 < 0 - 1

化简

- 2 cos (x) < - 1

- 2 cos (x) < - 1

在两边乘以

- 1

- 2 cos (x) < - 1

两边乘以 -1（不等式变号）

(- 2 cos (x)) (- 1) > (- 1) (- 1)

化简

2 cos (x) > 1

2 cos (x) > 1

两边除以

2

2 cos (x) > 1

两边除以

2

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

化简

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

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

1 - 2 cos (x) > 0 : cos (x) < \frac{1}{2}

1 - 2 cos (x) > 0

将

1

到右边

1 - 2 cos (x) > 0

两边减去

1

1 - 2 cos (x) - 1 > 0 - 1

化简

- 2 cos (x) > - 1

- 2 cos (x) > - 1

在两边乘以

- 1

- 2 cos (x) > - 1

两边乘以 -1（不等式变号）

(- 2 cos (x)) (- 1) < (- 1) (- 1)

化简

2 cos (x) < 1

2 cos (x) < 1

两边除以

2

2 cos (x) < 1

两边除以

2

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

化简

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

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

确定

cos (x)

符号

cos (x) = 0

cos (x) < 0

cos (x) > 0

找到奇点

找到分母的零解

cos (x) : cos (x) = 0

总结如下表：

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

确定满足所需条件的区间：

\leq 0

cos (x) < 0 or cos (x) = \frac{1}{2} or cos (x) > \frac{1}{2}

合并重叠的区间

cos (x) < 0 or cos (x) = \frac{1}{2} or cos (x) > \frac{1}{2}

两个区间的并集是指存在于任一区间的数的集合

cos (x) < 0

or

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

cos (x) < 0 or cos (x) = \frac{1}{2}

两个区间的并集是指存在于任一区间的数的集合

cos (x) < 0 or cos (x) = \frac{1}{2}

or

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

cos (x) < 0 or cos (x) \geq \frac{1}{2}

cos (x) < 0 or cos (x) \geq \frac{1}{2}

cos (x) < 0 or cos (x) \geq \frac{1}{2}

合并区间

(cos (x) \leq - \frac{1}{2} or cos (x) > 0) and (cos (x) < 0 or cos (x) \geq \frac{1}{2})

合并重叠的区间

cos (x) \leq - \frac{1}{2} or cos (x) > 0 and cos (x) < 0 or cos (x) \geq \frac{1}{2}

两个区间的交集是指同时存在于这两个区间的数的集合

cos (x) \leq - \frac{1}{2} or cos (x) > 0

and

cos (x) < 0 or cos (x) \geq \frac{1}{2}

cos (x) \leq - \frac{1}{2} or cos (x) \geq \frac{1}{2}

cos (x) \leq - \frac{1}{2} or cos (x) \geq \frac{1}{2}

cos (x) \leq - \frac{1}{2} : \frac{2 π}{3} + 2 πn \leq x \leq \frac{4 π}{3} + 2 πn

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

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (- \frac{1}{2}) + 2 πn \leq x \leq 2 π - arccos (- \frac{1}{2}) + 2 πn

化简

arccos (- \frac{1}{2}) : \frac{2 π}{3}

arccos (- \frac{1}{2})

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{2 π}{3}

化简

2 π - arccos (- \frac{1}{2}) : \frac{4 π}{3}

2 π - arccos (- \frac{1}{2})

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{2 π}{3}

化简

2 π - \frac{2 π}{3}

将项转换为分式:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{2 π}{3}

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{2 π 3 - 2 π}{3}

2 π 3 - 2 π = 4 π

2 π 3 - 2 π

数字相乘:

2 \cdot 3 = 6

= 6 π - 2 π

同类项相加:

6 π - 2 π = 4 π

= 4 π

= \frac{4 π}{3}

= \frac{4 π}{3}

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

cos (x) \geq \frac{1}{2} : - \frac{π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

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

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{1}{2}) + 2 πn \leq x \leq arccos (\frac{1}{2}) + 2 πn

化简

- arccos (\frac{1}{2}) : - \frac{π}{3}

- arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{3}

化简

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{3}

- \frac{π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

合并区间

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

合并重叠的区间

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

流行的例子

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

sin(θ)<0,cos(θ)>0

sin (θ) < 0, cos (θ) > 0

cot (x) < - 1

tan(θ)<0,cos(θ)>0

tan (θ) < 0, cos (θ) > 0

sin(x-45)> 1/2 sqrt(3)

sin (x - 4 5^{\circ}) > \frac{1}{2} 3