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受欢迎的三角函数 >

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

解答

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

解答

\frac{π}{12} + πn \leq x \leq \frac{11 π}{12} + πn

+2

间隔符号

[\frac{π}{12} + πn, \frac{11 π}{12} + πn]

十进制

0.26179 \dots + πn \leq x \leq 2.87979 \dots + πn

求解步骤

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

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (\frac{3}{2}) + 2 πn \leq 2 x \leq 2 π - arccos (\frac{3}{2}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arccos (\frac{3}{2}) + 2 πn \leq 2 x and 2 x \leq 2 π - arccos (\frac{3}{2}) + 2 πn

arccos (\frac{3}{2}) + 2 πn \leq 2 x : x \geq \frac{π}{12} + πn

arccos (\frac{3}{2}) + 2 πn \leq 2 x

交换两边

2 x \geq arccos (\frac{3}{2}) + 2 πn

化简

arccos (\frac{3}{2}) + 2 πn : \frac{π}{6} + 2 πn

arccos (\frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6} + 2 πn

2 x \geq \frac{π}{6} + 2 πn

两边除以

2

2 x \geq \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} \geq \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \geq \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x \geq \frac{π}{12} + πn

x \geq \frac{π}{12} + πn

x \geq \frac{π}{12} + πn

2 x \leq 2 π - arccos (\frac{3}{2}) + 2 πn : x \leq \frac{11 π}{12} + πn

2 x \leq 2 π - arccos (\frac{3}{2}) + 2 πn

化简

2 π - arccos (\frac{3}{2}) + 2 πn : 2 π - \frac{π}{6} + 2 πn

2 π - arccos (\frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{6} + 2 πn

2 x \leq 2 π - \frac{π}{6} + 2 πn

两边除以

2

2 x \leq 2 π - \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 π}{2} - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : π - \frac{π}{12} + πn

\frac{2 π}{2} - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= π - \frac{π}{12} + πn

x \leq π - \frac{π}{12} + πn

x \leq π - \frac{π}{12} + πn

化简

π - \frac{π}{12} : \frac{11 π}{12}

π - \frac{π}{12}

将项转换为分式:

π = \frac{π 12}{12}

= \frac{π 12}{12} - \frac{π}{12}

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{π 12 - π}{12}

同类项相加:

12 π - π = 11 π

= \frac{11 π}{12}

x \leq \frac{11 π}{12} + πn

x \leq \frac{11 π}{12} + πn

合并区间

x \geq \frac{π}{12} + πn and x \leq \frac{11 π}{12} + πn

合并重叠的区间

\frac{π}{12} + πn \leq x \leq \frac{11 π}{12} + πn

流行的例子

tan (x) < 3

sec (θ) > 0

pi/2-arctan(x)<0.0001

\frac{π}{2} - arctan (x) < 0.0001

2 cos (x) > 0

sin(x)> 1/(sqrt(2))

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