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受欢迎的三角函数 >

6cos(2x-60)<= 0

解答

6 cos (2 x - 60) \leq 0

解答

\frac{120 + π}{4} + πn \leq x \leq \frac{3 π + 120}{4} + πn

+2

间隔符号

[\frac{120 + π}{4} + πn, \frac{3 π + 120}{4} + πn]

十进制

30.78539 \dots + πn \leq x \leq 32.35619 \dots + πn

求解步骤

6 cos (2 x - 60) \leq 0

两边除以

6

6 cos (2 x - 60) \leq 0

两边除以

6

\frac{6 cos ( 2 x - 60 )}{6} \leq \frac{0}{6}

化简

cos (2 x - 60) \leq 0

cos (2 x - 60) \leq 0

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 (0) + 2 πn \leq (2 x - 60) \leq 2 π - arccos (0) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arccos (0) + 2 πn \leq 2 x - 60 and 2 x - 60 \leq 2 π - arccos (0) + 2 πn

arccos (0) + 2 πn \leq 2 x - 60 : x \geq \frac{120 + π}{4} + πn

arccos (0) + 2 πn \leq 2 x - 60

交换两边

2 x - 60 \geq arccos (0) + 2 πn

化简

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

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}{2} + 2 πn

2 x - 60 \geq \frac{π}{2} + 2 πn

将

60

到右边

2 x - 60 \geq \frac{π}{2} + 2 πn

两边加上

60

2 x - 60 + 60 \geq \frac{π}{2} + 2 πn + 60

化简

2 x \geq \frac{π}{2} + 2 πn + 60

2 x \geq \frac{π}{2} + 2 πn + 60

两边除以

2

2 x \geq \frac{π}{2} + 2 πn + 60

两边除以

2

\frac{2 x}{2} \geq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

化简

\frac{2 x}{2} \geq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{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}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2} : 30 + πn + \frac{π}{4}

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

对同类项分组

= \frac{60}{2} + \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

\frac{60}{2} = 30

\frac{60}{2}

数字相除:

\frac{60}{2} = 30

= 30

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

= 30 + πn + \frac{π}{4}

x \geq 30 + πn + \frac{π}{4}

x \geq 30 + πn + \frac{π}{4}

化简

30 + \frac{π}{4} : \frac{120 + π}{4}

30 + \frac{π}{4}

将项转换为分式:

30 = \frac{30 \cdot 4}{4}

= \frac{30 \cdot 4}{4} + \frac{π}{4}

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{30 \cdot 4 + π}{4}

数字相乘:

30 \cdot 4 = 120

= \frac{120 + π}{4}

x \geq \frac{120 + π}{4} + πn

x \geq \frac{120 + π}{4} + πn

2 x - 60 \leq 2 π - arccos (0) + 2 πn : x \leq \frac{3 π + 120}{4} + πn

2 x - 60 \leq 2 π - arccos (0) + 2 πn

化简

2 π - arccos (0) + 2 πn : 2 π - \frac{π}{2} + 2 πn

2 π - arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

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}{2} + 2 πn

2 x - 60 \leq 2 π - \frac{π}{2} + 2 πn

将

60

到右边

2 x - 60 \leq 2 π - \frac{π}{2} + 2 πn

两边加上

60

2 x - 60 + 60 \leq 2 π - \frac{π}{2} + 2 πn + 60

化简

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

两边除以

2

2 x \leq 2 π - \frac{π}{2} + 2 πn + 60

两边除以

2

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

化简

\frac{2 x}{2} \leq \frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{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}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2} : π + 30 + πn - \frac{π}{4}

\frac{2 π}{2} - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{60}{2}

对同类项分组

= \frac{2 π}{2} + \frac{60}{2} + \frac{2 πn}{2} - \frac{\frac{π}{2}}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{60}{2} = 30

\frac{60}{2}

数字相除:

\frac{60}{2} = 30

= 30

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

= π + 30 + πn - \frac{π}{4}

x \leq π + 30 + πn - \frac{π}{4}

x \leq π + 30 + πn - \frac{π}{4}

化简

π + 30 - \frac{π}{4} : \frac{3 π + 120}{4}

π + 30 - \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}, 30 = \frac{30 \cdot 4}{4}

= \frac{π 4}{4} + \frac{30 \cdot 4}{4} - \frac{π}{4}

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 4 + 30 \cdot 4 - π}{4}

π 4 + 30 \cdot 4 - π = 3 π + 120

π 4 + 30 \cdot 4 - π

对同类项分组

= 4 π - π + 30 \cdot 4

同类项相加:

4 π - π = 3 π

= 3 π + 30 \cdot 4

数字相乘:

30 \cdot 4 = 120

= 3 π + 120

= \frac{3 π + 120}{4}

x \leq \frac{3 π + 120}{4} + πn

x \leq \frac{3 π + 120}{4} + πn

合并区间

x \geq \frac{120 + π}{4} + πn and x \leq \frac{3 π + 120}{4} + πn

合并重叠的区间

\frac{120 + π}{4} + πn \leq x \leq \frac{3 π + 120}{4} + πn

流行的例子

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