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受欢迎的三角函数 >

(cos^2(x)-1/2)/(tan(x)-sqrt(3))<0

解答

\frac{cos ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

解答

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

+2

间隔符号

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

十进制

πn \leq x < 0.78539 \dots + πn or 1.04719 \dots + πn < x < 1.57079 \dots + πn or 2.35619 \dots + πn < x \leq 3.14159 \dots + πn

求解步骤

\frac{cos ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

利用以下特性:

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

因此

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

\frac{1 - sin ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

化简

\frac{1 - sin ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} : - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

\frac{1 - sin ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3}

乘以共轭根式

\frac{tan ( x ) + 3}{tan ( x ) + 3}

= \frac{( 1 - sin ^{2} ( x ) - \frac{1}{2} ) ( tan ( x ) + 3 )}{( tan ( x ) - 3 ) ( tan ( x ) + 3 )}

化简

(1 - sin^{2} (x) - \frac{1}{2}) (tan (x) + 3) : (- sin^{2} (x) + \frac{1}{2}) (tan (x) + 3)

(1 - sin^{2} (x) - \frac{1}{2}) (tan (x) + 3)

化简

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

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

合并分式

1 - \frac{1}{2} : \frac{1}{2}

1 - \frac{1}{2}

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= \frac{1 \cdot 2}{2} - \frac{1}{2}

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 \cdot 2 - 1}{2}

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

数字相乘:

1 \cdot 2 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

= - sin^{2} (x)

= (- sin^{2} (x) + \frac{1}{2}) (tan (x) + 3)

(tan (x) - 3) (tan (x) + 3) = tan^{2} (x) - 3

(tan (x) - 3) (tan (x) + 3)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = tan (x), b = 3

= tan^{2} (x) - (3)^{2}

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 3

= tan^{2} (x) - 3

= \frac{( - sin ^{2} ( x ) + \frac{1}{2} ) ( tan ( x ) + 3 )}{tan ^{2} ( x ) - 3}

分解

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

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

因式分解出通项

- 1

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

分解

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

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

使用根式运算法则:

a = (a)^{2}

\frac{1}{2} = (\frac{1}{2})^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} ) ( tan ( x ) + 3 )}{tan ^{2} ( x ) - 3}

分解

tan^{2} (x) - 3 : (tan (x) + 3) (tan (x) - 3)

tan^{2} (x) - 3

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

= tan^{2} (x) - (3)^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - (3)^{2} = (tan (x) + 3) (tan (x) - 3)

= (tan (x) + 3) (tan (x) - 3)

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} ) ( tan ( x ) + 3 )}{( tan ( x ) + 3 ) ( tan ( x ) - 3 )}

约分:

tan (x) + 3

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3} < 0

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

的周期:

π

\frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

包含以下函数及对应周期:

sin (x)

的周期为

2 π

复合周期为:

= π

用 sin, cos 表示

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3} < 0

使用基本三角恒等式:

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

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} < 0

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} < 0

化简

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3} : - \frac{cos ( x ) ( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{sin ( x ) - 3 cos ( x )}

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x )}{c o s ( x )} - 3}

化简

\frac{sin ( x )}{cos ( x )} - 3 : \frac{sin ( x ) - 3 cos ( x )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} - 3

将项转换为分式:

3 = \frac{3 cos ( x )}{cos ( x )}

= \frac{sin ( x )}{cos ( x )} - \frac{3 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 ) - 3 cos ( x )}{cos ( x )}

= - \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{\frac{s i n ( x ) - 3 c o s ( x )}{c o s ( x )}}

使用分式法则:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

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

- \frac{cos ( x ) ( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{sin ( x ) - 3 cos ( x )} < 0

确定

0 \leq x < π

时

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

- \frac{cos ( x ) ( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{sin ( x ) - 3 cos ( x )} = 0, 0 \leq x < π : x = \frac{π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

- \frac{cos ( x ) ( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{sin ( x ) - 3 cos ( x )} = 0, 0 \leq x < π

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

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

分别求解每个部分

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

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

cos (x) = 0, 0 \leq x < π

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 < π

范围内的解

x = \frac{π}{2}

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

无解

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

将

\frac{1}{2}

到右边

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

两边减去

\frac{1}{2}

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

化简

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

使用反三角函数性质

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{1}{2}) + 2 πn, x = π + arcsin (\frac{1}{2}) + 2 πn

x = arcsin (- \frac{1}{2}) + 2 πn, x = π + arcsin (\frac{1}{2}) + 2 πn

在

0 \leq x < π

范围内的解

无解

sin (x) - \frac{1}{2} = 0, 0 \leq x < π : x = \frac{π}{4}, x = \frac{3 π}{4}

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

将

\frac{1}{2}

到右边

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

两边加上

\frac{1}{2}

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

化简

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

使用反三角函数性质

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{2}) + 2 πn, x = π - arcsin (\frac{1}{2}) + 2 πn

x = arcsin (\frac{1}{2}) + 2 πn, x = π - arcsin (\frac{1}{2}) + 2 πn

在

0 \leq x < π

范围内的解

x = \frac{π}{4}, x = \frac{3 π}{4}

合并所有解

x = \frac{π}{2}, x = \frac{π}{4}, x = \frac{3 π}{4}

确定无定义点:

x = \frac{π}{3}

找到分母的零解

sin (x) - 3 cos (x) = 0

使用三角恒等式改写

sin (x) - 3 cos (x) = 0

在两边除以

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

\frac{sin ( x ) - 3 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

\frac{sin ( x )}{cos ( x )} - 3 = 0

使用基本三角恒等式:

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

tan (x) - 3 = 0

tan (x) - 3 = 0

将

3

到右边

tan (x) - 3 = 0

两边加上

3

tan (x) - 3 + 3 = 0 + 3

化简

tan (x) = 3

tan (x) = 3

tan (x) = 3

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{π}{3} + πn

x = \frac{π}{3} + πn

在

0 \leq x < π

范围内的解

x = \frac{π}{3}

\frac{π}{4}, \frac{π}{3}, \frac{π}{2}, \frac{3 π}{4}

确定区间

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

总结如下表：

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

确定满足所需条件的区间：

< 0

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

合并重叠的区间

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

0 < x < \frac{π}{4}

0 \leq x < \frac{π}{4}

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}{4}

or

\frac{π}{3} < x < \frac{π}{2}

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

or

\frac{3 π}{4} < x < π

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

or

x = π

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

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

使用周期

- \frac{( sin ( x ) + \frac{1}{2} ) ( sin ( x ) - \frac{1}{2} )}{tan ( x ) - 3}

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

流行的例子

tan(θ)>0,cot(θ)>0

tan (θ) > 0, cot (θ) > 0

6 sin (θ) \geq 0

tan(x)>\sqrt[4]{5},-pi<= x<= pi

tan (x) > 45, - π \leq x \leq π

2sin^2(x)+3sin(x)>= 2,0<= x<= 2pi

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

sin(x)+cos(x)>1

sin (x) + cos (x) > 1