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受欢迎的三角函数 >

sec(x)<= cos(x)

解答

sec (x) \leq cos (x)

解答

\frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

+2

间隔符号

(\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn)

十进制

1.57079 \dots + 2 πn < x < 4.71238 \dots + 2 πn

求解步骤

sec (x) \leq cos (x)

将

cos (x)

para o lado esquerdo

sec (x) \leq cos (x)

两边减去

cos (x)

sec (x) - cos (x) \leq cos (x) - cos (x)

sec (x) - cos (x) \leq 0

sec (x) - cos (x) \leq 0

sec (x) - cos (x)

的周期:

2 π

周期函数和的复合周期是这些周期的最小公倍数

sec (x), cos (x)

sec (x)

的周期:

2 π

sec (x)

的周期是

2 π

= 2 π

cos (x)

的周期:

2 π

cos (x)

的周期是

2 π

= 2 π

合并周期:

2 π, 2 π

= 2 π

用 sin, cos 表示

sec (x) - cos (x) \leq 0

使用基本三角恒等式:

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

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

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

化简

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

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

将项转换为分式:

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

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

1 - cos (x) cos (x) = 1 - cos^{2} (x)

1 - cos (x) cos (x)

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

= 1 - cos^{2} (x)

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

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

确定

0 \leq x < 2 π

时

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

的零点和无定义点

要找到零点，将不等式设置为零

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

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

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

用替代法求解

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

令

: cos (x) = u

\frac{1 - u ^{2}}{u} = 0

\frac{1 - u ^{2}}{u} = 0 : u = 1, u = - 1

\frac{1 - u ^{2}}{u} = 0

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

1 - u^{2} = 0

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1 - u ^{2}}{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 = - 1

u = cos (x)

代回

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1, 0 \leq x < 2 π : x = 0

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 = 0 + 2 πn

x = 0 + 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

在

0 \leq x < 2 π

范围内的解

x = 0

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 = 0, 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}

0, \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：

1 - cos^{2} (x) cos (x) \frac{1 - c o s ^{2} ( x )}{c o s ( x )} x = 0 0 + 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 0 + 0

确定满足所需条件的区间：

\leq 0

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

合并重叠的区间

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

\frac{π}{2} < x < π

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

or

x = π

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

or

π < x < \frac{3 π}{2}

x = 0 or \frac{π}{2} < 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

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

or

x = 2 π

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

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

使用周期

sec (x) - cos (x)

\frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

流行的例子

- sin (2 x) > 0

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

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

6-3cos((pit)/2)>6

6 - 3 cos (\frac{π t}{2}) > 6

(10pi)/9 <= arctan(θ)

\frac{10 π}{9} \leq arctan (θ)

cos (x) - 1 \geq - 2