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受欢迎的三角函数 >

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

解答

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

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

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

利用以下特性:

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

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

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

的周期:

2 π

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

的零点和无定义点

要找到零点，将不等式设置为零

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

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

无解

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

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

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}

因为方程对以下值无定义:

\frac{π}{2}, \frac{3 π}{2}

x \in R 无解

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}

找到分母的零解

1 + cos^{2} (x) - sin^{2} (x) = 0

使用三角恒等式改写

1 + cos^{2} (x) - sin^{2} (x)

使用倍角公式:

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

= 1 + cos (2 x)

1 + cos (2 x) = 0

将

1

到右边

1 + cos (2 x) = 0

两边减去

1

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

化简

cos (2 x) = - 1

cos (2 x) = - 1

cos (2 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}

2 x = π + 2 πn

2 x = π + 2 πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

\frac{π}{2}, \frac{3 π}{2}

确定区间

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

总结如下表：

cos (x) 1 + cos^{2} (x) - sin^{2} (x) \frac{c o s ( x )}{1 + c o s ^{2} ( x ) - s i n ^{2} ( x )} x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} 00 未定义 \frac{π}{2} < x < \frac{3 π}{2} - + - x = \frac{3 π}{2} 00 未定义 \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

确定满足所需条件的区间：

< 0

\frac{π}{2} < x < \frac{3 π}{2}

使用周期

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

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

流行的例子

(tan(x)-tan^2(x))/(2sin(x)-1)<0

\frac{tan ( x ) - tan ^{2} ( x )}{2 sin ( x ) - 1} < 0

sin(x)-cos(x)>1

sin (x) - cos (x) > 1

sin (y) > 0

2 sin (x) + 1 \geq 0

cos^{2} (x) > \frac{5}{6}