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受欢迎的三角函数 >

tan(t)-tan^2(t)+sec^3(t)>0

解答

tan (t) - tan^{2} (t) + sec^{3} (t) > 0

解答

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

+2

间隔符号

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

十进制

2 πn \leq t < 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn < t \leq 6.28318 \dots + 2 πn

求解步骤

tan (t) - tan^{2} (t) + sec^{3} (t) > 0

tan (t) - tan^{2} (t) + sec^{3} (t)

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

tan (t), tan^{2} (t), sec^{3} (t)

tan (t)

的周期:

π

tan (x)

的周期是

π

= π

tan^{2} (t)

的周期:

π

tan^{n} (x)

的周期

= tan (x)

的周期

tan (t)

的周期:

π

tan (x)

的周期是

π

= π

π

sec^{3} (t)

的周期:

2 π

sec^{n} (x)

的周期

= sec (x)

的周期

,

若 n 为奇数

sec (t)

的周期:

2 π

sec (x)

的周期是

2 π

= 2 π

2 π

合并周期:

π, π, 2 π

= 2 π

用 sin, cos 表示

tan (t) - tan^{2} (t) + sec^{3} (t) > 0

使用基本三角恒等式:

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

\frac{sin ( t )}{cos ( t )} - (\frac{sin ( t )}{cos ( t )})^{2} + sec^{3} (t) > 0

使用基本三角恒等式:

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

\frac{sin ( t )}{cos ( t )} - (\frac{sin ( t )}{cos ( t )})^{2} + (\frac{1}{cos ( t )})^{3} > 0

\frac{sin ( t )}{cos ( t )} - (\frac{sin ( t )}{cos ( t )})^{2} + (\frac{1}{cos ( t )})^{3} > 0

化简

\frac{sin ( t )}{cos ( t )} - (\frac{sin ( t )}{cos ( t )})^{2} + (\frac{1}{cos ( t )})^{3} : \frac{cos ^{2} ( t ) sin ( t ) - sin ^{2} ( t ) cos ( t ) + 1}{cos ^{3} ( t )}

\frac{sin ( t )}{cos ( t )} - (\frac{sin ( t )}{cos ( t )})^{2} + (\frac{1}{cos ( t )})^{3}

(\frac{sin ( t )}{cos ( t )})^{2} = \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

(\frac{sin ( t )}{cos ( t )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

(\frac{1}{cos ( t )})^{3} = \frac{1}{cos ^{3} ( t )}

(\frac{1}{cos ( t )})^{3}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{3}}{cos ^{3} ( t )}

使用法则

1^{a} = 1

1^{3} = 1

= \frac{1}{cos ^{3} ( t )}

= \frac{sin ( t )}{cos ( t )} - \frac{sin ^{2} ( t )}{cos ^{2} ( t )} + \frac{1}{cos ^{3} ( t )}

cos (t), cos^{2} (t), cos^{3} (t)

的最小公倍数:

cos^{3} (t)

cos (t), cos^{2} (t), cos^{3} (t)

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= cos^{3} (t)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos^{3} (t)

对于

\frac{sin ( t )}{cos ( t )} :

将分母和分子乘以

cos^{2} (t)

\frac{sin ( t )}{cos ( t )} = \frac{sin ( t ) cos ^{2} ( t )}{cos ( t ) cos ^{2} ( t )} = \frac{sin ( t ) cos ^{2} ( t )}{cos ^{3} ( t )}

对于

\frac{sin ^{2} ( t )}{cos ^{2} ( t )} :

将分母和分子乘以

cos (t)

\frac{sin ^{2} ( t )}{cos ^{2} ( t )} = \frac{sin ^{2} ( t ) cos ( t )}{cos ^{2} ( t ) cos ( t )} = \frac{sin ^{2} ( t ) cos ( t )}{cos ^{3} ( t )}

= \frac{sin ( t ) cos ^{2} ( t )}{cos ^{3} ( t )} - \frac{sin ^{2} ( t ) cos ( t )}{cos ^{3} ( t )} + \frac{1}{cos ^{3} ( t )}

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{sin ( t ) cos ^{2} ( t ) - sin ^{2} ( t ) cos ( t ) + 1}{cos ^{3} ( t )}

\frac{cos ^{2} ( t ) sin ( t ) - sin ^{2} ( t ) cos ( t ) + 1}{cos ^{3} ( t )} > 0

确定

0 \leq t < 2 π

时

\frac{cos ^{2} ( t ) sin ( t ) - sin ^{2} ( t ) cos ( t ) + 1}{cos ^{3} ( t )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{cos ^{2} ( t ) sin ( t ) - sin ^{2} ( t ) cos ( t ) + 1}{cos ^{3} ( t )} = 0

确定无定义点:

t = \frac{π}{2}, t = \frac{3 π}{2}

找到分母的零解

cos^{3} (t) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

cos (t) = 0

cos (t) = 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}

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

在

0 \leq t < 2 π

范围内的解

t = \frac{π}{2}, t = \frac{3 π}{2}

\frac{π}{2}, \frac{3 π}{2}

确定区间

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

总结如下表：

cos^{2} (t) sin (t) - sin^{2} (t) cos (t) + 1 cos^{3} (t) \frac{c o s ^{2} ( t ) s i n ( t ) - s i n ^{2} ( t ) c o s ( t ) + 1}{c o s ^{3} ( t )} t = 0 + + + 0 < t < \frac{π}{2} + + + t = \frac{π}{2} + 0 未定义 \frac{π}{2} < t < \frac{3 π}{2} + - - t = \frac{3 π}{2} + 0 未定义 \frac{3 π}{2} < t < 2 π + + + t = 2 π + + +

确定满足所需条件的区间：

> 0

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

合并重叠的区间

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

两个区间的并集是指存在于任一区间的数的集合

t = 0

or

0 < t < \frac{π}{2}

0 \leq t < \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 t < \frac{π}{2}

or

\frac{3 π}{2} < t < 2 π

0 \leq t < \frac{π}{2} or \frac{3 π}{2} < t < 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 t < \frac{π}{2} or \frac{3 π}{2} < t < 2 π

or

t = 2 π

0 \leq t < \frac{π}{2} or \frac{3 π}{2} < t \leq 2 π

0 \leq t < \frac{π}{2} or \frac{3 π}{2} < t \leq 2 π

使用周期

tan (t) - tan^{2} (t) + sec^{3} (t)

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

流行的例子

-cos(2x)<= (sqrt(3))/2

- cos (2 x) \leq \frac{3}{2}

sin(x)<0,sec(x)>0

sin (x) < 0, sec (x) > 0

2sin(x)+3((sin(2x))/(2sin(x)))<0

2 sin (x) + 3 (\frac{sin ( 2 x )}{2 sin ( x )}) < 0

cos^2(x)>sin(x)cos(x)

cos^{2} (x) > sin (x) cos (x)

cos(θ)>0,sin(θ)>0

cos (θ) > 0, sin (θ) > 0