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受欢迎的三角函数 >

((1-cos(x))(1+cos(x)))/(sin(x)+cos(x))>0

解答

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

解答

2 πn < x < \frac{3 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

+2

间隔符号

(2 πn, \frac{3 π}{4} + 2 πn) \cup (\frac{7 π}{4} + 2 πn, 2 π + 2 πn)

十进制

2 πn < x < 2.35619 \dots + 2 πn or 5.49778 \dots + 2 πn < x < 6.28318 \dots + 2 πn

求解步骤

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

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

的周期:

2 π

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

包含以下函数及对应周期:

cos (x)

的周期为

2 π

复合周期为:

= 2 π

确定

0 \leq x < 2 π

时

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

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

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

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

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

分别求解每个部分

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

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

1 - cos (x) = 0, 0 \leq x < 2 π

将

1

到右边

1 - cos (x) = 0

两边减去

1

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

化简

- cos (x) = - 1

- cos (x) = - 1

两边除以

- 1

- cos (x) = - 1

两边除以

- 1

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

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

1 + cos (x) = 0, 0 \leq x < 2 π

将

1

到右边

1 + cos (x) = 0

两边减去

1

1 + cos (x) - 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 = 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{3 π}{4}, x = \frac{7 π}{4}

找到分母的零解

sin (x) + cos (x) = 0

使用三角恒等式改写

sin (x) + cos (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

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

化简

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

使用基本三角恒等式:

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

tan (x) + 1 = 0

tan (x) + 1 = 0

将

1

到右边

tan (x) + 1 = 0

两边减去

1

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

化简

tan (x) = - 1

tan (x) = - 1

tan (x) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

在

0 \leq x < 2 π

范围内的解

x = \frac{3 π}{4}, x = \frac{7 π}{4}

0, \frac{3 π}{4}, π, \frac{7 π}{4}

确定区间

0 < x < \frac{3 π}{4}, \frac{3 π}{4} < x < π, π < x < \frac{7 π}{4}, \frac{7 π}{4} < x < 2 π

总结如下表：

1 - cos (x) 1 + cos (x) sin (x) + cos (x) \frac{( 1 - c o s ( x ) ) ( 1 + c o s ( x ) )}{s i n ( x ) + c o s ( x )} x = 0 0 + + 0 0 < x < \frac{3 π}{4} + + + + x = \frac{3 π}{4} + + 0 未定义 \frac{3 π}{4} < x < π + + - - x = π + 0 - 0 π < x < \frac{7 π}{4} + + - - x = \frac{7 π}{4} + + 0 未定义 \frac{7 π}{4} < x < 2 π + + + + x = 2 π 0 + + 0

确定满足所需条件的区间：

> 0

0 < x < \frac{3 π}{4} or \frac{7 π}{4} < x < 2 π

使用周期

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

2 πn < x < \frac{3 π}{4} + 2 πn or \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

流行的例子

(2sin(x)+1)(-2cos(x)+sqrt(3))>0

(2 sin (x) + 1) (- 2 cos (x) + 3) > 0

cos (x) > - sin (x)

2sin^2(x)+sqrt(3)sin(x)>= 0

2 sin^{2} (x) + 3 sin (x) \geq 0

arcsin((sqrt(3))/2-(0.15)/x)>=-pi/2

arcsin (\frac{3}{2} - \frac{0.15}{x}) \geq - \frac{π}{2}

cos(x)(2sin(x)-sqrt(3))>= 0

cos (x) (2 sin (x) - 3) \geq 0