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受欢迎的三角函数 >

tan(x)>= sin(x)

解答

tan (x) \geq sin (x)

解答

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

+2

间隔符号

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

十进制

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

求解步骤

tan (x) \geq sin (x)

将

sin (x)

para o lado esquerdo

tan (x) \geq sin (x)

两边减去

sin (x)

tan (x) - sin (x) \geq sin (x) - sin (x)

tan (x) - sin (x) \geq 0

tan (x) - sin (x) \geq 0

tan (x) - sin (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

tan (x), sin (x)

tan (x)

的周期:

π

tan (x)

的周期是

π

= π

sin (x)

的周期:

2 π

sin (x)

的周期是

2 π

= 2 π

合并周期:

π, 2 π

= 2 π

用 sin, cos 表示

tan (x) - sin (x) \geq 0

使用基本三角恒等式:

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

\frac{sin ( x )}{cos ( x )} - sin (x) \geq 0

\frac{sin ( x )}{cos ( x )} - sin (x) \geq 0

化简

\frac{sin ( x )}{cos ( x )} - sin (x) : \frac{sin ( x ) - sin ( x ) cos ( x )}{cos ( x )}

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

将项转换为分式:

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

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

\frac{sin ( x ) - sin ( x ) cos ( x )}{cos ( x )} \geq 0

确定

0 \leq x < 2 π

时

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

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

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

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

sin (x) - sin (x) cos (x) = 0

分解

sin (x) - sin (x) cos (x) : - sin (x) (cos (x) - 1)

sin (x) - sin (x) cos (x)

因式分解出通项

- sin (x)

= - sin (x) (- 1 + cos (x))

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

分别求解每个部分

sin (x) = 0 or cos (x) - 1 = 0

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

cos (x) - 1 = 0, 0 \leq x < 2 π : x = 0

cos (x) - 1 = 0, 0 \leq x < 2 π

将

1

到右边

cos (x) - 1 = 0

两边加上

1

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

化简

cos (x) = 1

cos (x) = 1

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

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：

sin (x) - sin (x) cos (x) cos (x) \frac{s i n ( x ) - s i n ( x ) c o s ( 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：

\geq 0

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

合并重叠的区间

0 \leq x < \frac{π}{2} or π \leq 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

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

π < x < \frac{3 π}{2}

0 \leq x < \frac{π}{2} or π \leq 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

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

or

x = 2 π

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

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

使用周期

tan (x) - sin (x)

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

流行的例子

1 \leq tan (x)

sin (9 x^{2}) > 0

2 cos (2 x) - 1 > 0

0 < cos (x)

9^{1+sin^2(pix)}+30*9^{cos^2(pix)}<= 117

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{c o s^{2} (π x)} \leq 117