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受欢迎的三角函数 >

2cos(4x-pi/3)-1<= 0

解答

2 cos (4 x - \frac{π}{3}) - 1 \leq 0

解答

\frac{π}{6} + \frac{π}{2} n \leq x \leq \frac{π}{2} + \frac{π}{2} n

+2

间隔符号

[\frac{π}{6} + \frac{π}{2} n, \frac{π}{2} + \frac{π}{2} n]

十进制

0.52359 \dots + \frac{π}{2} n \leq x \leq 1.57079 \dots + \frac{π}{2} n

求解步骤

2 cos (4 x - \frac{π}{3}) - 1 \leq 0

将

1

到右边

2 cos (4 x - \frac{π}{3}) - 1 \leq 0

两边加上

1

2 cos (4 x - \frac{π}{3}) - 1 + 1 \leq 0 + 1

化简

2 cos (4 x - \frac{π}{3}) \leq 1

2 cos (4 x - \frac{π}{3}) \leq 1

两边除以

2

2 cos (4 x - \frac{π}{3}) \leq 1

两边除以

2

\frac{2 cos ( 4 x - \frac{π}{3} )}{2} \leq \frac{1}{2}

化简

cos (4 x - \frac{π}{3}) \leq \frac{1}{2}

cos (4 x - \frac{π}{3}) \leq \frac{1}{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{1}{2}) + 2 πn \leq (4 x - \frac{π}{3}) \leq 2 π - arccos (\frac{1}{2}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arccos (\frac{1}{2}) + 2 πn \leq 4 x - \frac{π}{3} and 4 x - \frac{π}{3} \leq 2 π - arccos (\frac{1}{2}) + 2 πn

arccos (\frac{1}{2}) + 2 πn \leq 4 x - \frac{π}{3} : x \geq \frac{π}{6} + \frac{πn}{2}

arccos (\frac{1}{2}) + 2 πn \leq 4 x - \frac{π}{3}

交换两边

4 x - \frac{π}{3} \geq arccos (\frac{1}{2}) + 2 πn

化简

arccos (\frac{1}{2}) + 2 πn : \frac{π}{3} + 2 πn

arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

4 x - \frac{π}{3} \geq \frac{π}{3} + 2 πn

将

\frac{π}{3}

到右边

4 x - \frac{π}{3} \geq \frac{π}{3} + 2 πn

两边加上

\frac{π}{3}

4 x - \frac{π}{3} + \frac{π}{3} \geq \frac{π}{3} + 2 πn + \frac{π}{3}

化简

4 x - \frac{π}{3} + \frac{π}{3} \geq \frac{π}{3} + 2 πn + \frac{π}{3}

化简

4 x - \frac{π}{3} + \frac{π}{3} : 4 x

4 x - \frac{π}{3} + \frac{π}{3}

同类项相加:

- \frac{π}{3} + \frac{π}{3} \geq 0

= 4 x

化简

\frac{π}{3} + 2 πn + \frac{π}{3} : \frac{2 π}{3} + 2 πn

\frac{π}{3} + 2 πn + \frac{π}{3}

对同类项分组

= \frac{π}{3} + \frac{π}{3} + 2 πn

合并分式

\frac{π}{3} + \frac{π}{3} : \frac{2 π}{3}

使用法则

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

= \frac{π + π}{3}

同类项相加:

π + π = 2 π

= \frac{2 π}{3}

= \frac{2 π}{3} + 2 πn

4 x \geq \frac{2 π}{3} + 2 πn

4 x \geq \frac{2 π}{3} + 2 πn

4 x \geq \frac{2 π}{3} + 2 πn

两边除以

4

4 x \geq \frac{2 π}{3} + 2 πn

两边除以

4

\frac{4 x}{4} \geq \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \geq \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

使用分式法则:

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

= \frac{2 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{2 π}{12}

约分:

2

= \frac{π}{6}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{6} + \frac{πn}{2}

x \geq \frac{π}{6} + \frac{πn}{2}

x \geq \frac{π}{6} + \frac{πn}{2}

x \geq \frac{π}{6} + \frac{πn}{2}

4 x - \frac{π}{3} \leq 2 π - arccos (\frac{1}{2}) + 2 πn : x \leq \frac{π}{2} + \frac{πn}{2}

4 x - \frac{π}{3} \leq 2 π - arccos (\frac{1}{2}) + 2 πn

化简

2 π - arccos (\frac{1}{2}) + 2 πn : 2 π - \frac{π}{3} + 2 πn

2 π - arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

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

将

\frac{π}{3}

到右边

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

两边加上

\frac{π}{3}

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

化简

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

化简

4 x - \frac{π}{3} + \frac{π}{3} : 4 x

4 x - \frac{π}{3} + \frac{π}{3}

同类项相加:

- \frac{π}{3} + \frac{π}{3} \leq 0

= 4 x

化简

2 π - \frac{π}{3} + 2 πn + \frac{π}{3} : 2 π + 2 πn

2 π - \frac{π}{3} + 2 πn + \frac{π}{3}

对同类项分组

= - \frac{π}{3} + \frac{π}{3} + 2 π + 2 πn

同类项相加:

- \frac{π}{3} + \frac{π}{3} = 0

= 2 π + 2 πn

4 x \leq 2 π + 2 πn

4 x \leq 2 π + 2 πn

4 x \leq 2 π + 2 πn

两边除以

4

4 x \leq 2 π + 2 πn

两边除以

4

\frac{4 x}{4} \leq \frac{2 π}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \leq \frac{2 π}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{2 π}{4} + \frac{2 πn}{4} : \frac{π}{2} + \frac{πn}{2}

\frac{2 π}{4} + \frac{2 πn}{4}

消掉

\frac{2 π}{4} : \frac{π}{2}

\frac{2 π}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{4}

消掉

\frac{2 πn}{4} : \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{2} + \frac{πn}{2}

x \leq \frac{π}{2} + \frac{πn}{2}

x \leq \frac{π}{2} + \frac{πn}{2}

x \leq \frac{π}{2} + \frac{πn}{2}

合并区间

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

合并重叠的区间

\frac{π}{6} + \frac{π}{2} n \leq x \leq \frac{π}{2} + \frac{π}{2} n

流行的例子

sin (x) \leq - \frac{5}{2}

arccos((x-5)/(10))-pi/3 >= 0

arccos (\frac{x - 5}{10}) - \frac{π}{3} \geq 0

tan(3x-1)>-sqrt(3)

tan (3 x - 1) > - 3

sin(x)>= 0,(1-2sin(x))/(1+sin(x))>0

sin (x) \geq 0, \frac{1 - 2 sin ( x )}{1 + sin ( x )} > 0

4cos(x/2)+1>2sqrt(2)+1,0<= x<= 6pi

4 cos (\frac{x}{2}) + 1 > 22 + 1, 0 \leq x \leq 6 π