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受欢迎的三角函数 >

5<= 20cos(pi/(20)(x-20))+23<= 20

解答

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20

解答

\frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n or \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

+2

间隔符号

[\frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n, \frac{20 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n] \cup [\frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n, \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n]

十进制

2.87132 \dots + 40 n \leq x \leq 9.04145 \dots + 40 n or 30.95854 \dots + 40 n \leq x \leq 37.12867 \dots + 40 n

求解步骤

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20

若

a \leq u \leq b

，则

a \leq u and u \leq b

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23 and 20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23 : \frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

5 \leq 20 cos (\frac{π}{20} (x - 20)) + 23

交换两边

20 cos (\frac{π}{20} (x - 20)) + 23 \geq 5

将

23

到右边

20 cos (\frac{π}{20} (x - 20)) + 23 \geq 5

两边减去

23

20 cos (\frac{π}{20} (x - 20)) + 23 - 23 \geq 5 - 23

化简

20 cos (\frac{π}{20} (x - 20)) \geq - 18

20 cos (\frac{π}{20} (x - 20)) \geq - 18

两边除以

20

20 cos (\frac{π}{20} (x - 20)) \geq - 18

两边除以

20

\frac{20 cos ( \frac{π}{20} ( x - 20 ) )}{20} \geq \frac{- 18}{20}

化简

\frac{20 cos ( \frac{π}{20} ( x - 20 ) )}{20} \geq \frac{- 18}{20}

化简

\frac{20 cos ( \frac{π}{20} ( x - 20 ) )}{20} : cos (\frac{π}{20} (x - 20))

\frac{20 cos ( \frac{π}{20} ( x - 20 ) )}{20}

数字相除:

\frac{20}{20} = 1

= cos (\frac{π}{20} (x - 20))

化简

\frac{- 18}{20} : - \frac{9}{10}

\frac{- 18}{20}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{18}{20}

约分:

2

= - \frac{9}{10}

cos (\frac{π}{20} (x - 20)) \geq - \frac{9}{10}

cos (\frac{π}{20} (x - 20)) \geq - \frac{9}{10}

cos (\frac{π}{20} (x - 20)) \geq - \frac{9}{10}

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (- \frac{9}{10}) + 2 πn \leq \frac{π}{20} (x - 20) \leq arccos (- \frac{9}{10}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (- \frac{9}{10}) + 2 πn \leq \frac{π}{20} (x - 20) and \frac{π}{20} (x - 20) \leq arccos (- \frac{9}{10}) + 2 πn

- arccos (- \frac{9}{10}) + 2 πn \leq \frac{π}{20} (x - 20) : x \geq \frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

- arccos (- \frac{9}{10}) + 2 πn \leq \frac{π}{20} (x - 20)

交换两边

\frac{π}{20} (x - 20) \geq - arccos (- \frac{9}{10}) + 2 πn

在两边乘以

20

\frac{π}{20} (x - 20) \geq - arccos (- \frac{9}{10}) + 2 πn

在两边乘以

20

20 \cdot \frac{π}{20} (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) : π (x - 20)

20 \cdot \frac{π}{20} (x - 20)

分式相乘:

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

= \frac{20 π}{20} (x - 20)

约分:

20

= (x - 20) π

化简

- 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn : - 20 arccos (- \frac{9}{10}) + 40 πn

- 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

数字相乘:

20 \cdot 2 = 40

= - 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 40 πn

两边除以

π

π (x - 20) \geq - 20 arccos (- \frac{9}{10}) + 40 πn

两边除以

π

\frac{π ( x - 20 )}{π} \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + \frac{40 πn}{π}

化简

x - 20 \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

x - 20 \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

将

20

到右边

x - 20 \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

两边加上

20

x - 20 + 20 \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

化简

x \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

x \geq - \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

化简

- \frac{20 arccos ( - \frac{9}{10} )}{π} + 20 : \frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π}

- \frac{20 arccos ( - \frac{9}{10} )}{π} + 20

将项转换为分式:

20 = \frac{20 π}{π}

= - \frac{20 arccos ( - \frac{9}{10} )}{π} + \frac{20 π}{π}

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{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π}

x \geq \frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

\frac{π}{20} (x - 20) \leq arccos (- \frac{9}{10}) + 2 πn : x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

\frac{π}{20} (x - 20) \leq arccos (- \frac{9}{10}) + 2 πn

在两边乘以

20

\frac{π}{20} (x - 20) \leq arccos (- \frac{9}{10}) + 2 πn

在两边乘以

20

20 \cdot \frac{π}{20} (x - 20) \leq 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) \leq 20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) : π (x - 20)

20 \cdot \frac{π}{20} (x - 20)

分式相乘:

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

= \frac{20 π}{20} (x - 20)

约分:

20

= (x - 20) π

化简

20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn : 20 arccos (- \frac{9}{10}) + 40 πn

20 arccos (- \frac{9}{10}) + 20 \cdot 2 πn

数字相乘:

20 \cdot 2 = 40

= 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \leq 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \leq 20 arccos (- \frac{9}{10}) + 40 πn

π (x - 20) \leq 20 arccos (- \frac{9}{10}) + 40 πn

两边除以

π

π (x - 20) \leq 20 arccos (- \frac{9}{10}) + 40 πn

两边除以

π

\frac{π ( x - 20 )}{π} \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + \frac{40 πn}{π}

化简

x - 20 \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

x - 20 \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

将

20

到右边

x - 20 \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n

两边加上

20

x - 20 + 20 \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

化简

x \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

x \leq \frac{20 arccos ( - \frac{9}{10} )}{π} + 40 n + 20

化简

\frac{20 arccos ( - \frac{9}{10} )}{π} + 20 : \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π}

\frac{20 arccos ( - \frac{9}{10} )}{π} + 20

将项转换为分式:

20 = \frac{20 π}{π}

= \frac{20 arccos ( - \frac{9}{10} )}{π} + \frac{20 π}{π}

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{20 arccos ( - \frac{9}{10} ) + 20 π}{π}

x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

合并区间

x \geq \frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n and x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

合并重叠的区间

\frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20 : \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n \leq x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20

将

23

到右边

20 cos (\frac{π}{20} (x - 20)) + 23 \leq 20

两边减去

23

20 cos (\frac{π}{20} (x - 20)) + 23 - 23 \leq 20 - 23

化简

20 cos (\frac{π}{20} (x - 20)) \leq - 3

20 cos (\frac{π}{20} (x - 20)) \leq - 3

两边除以

20

20 cos (\frac{π}{20} (x - 20)) \leq - 3

两边除以

20

\frac{20 cos ( \frac{π}{20} ( x - 20 ) )}{20} \leq \frac{- 3}{20}

化简

cos (\frac{π}{20} (x - 20)) \leq - \frac{3}{20}

cos (\frac{π}{20} (x - 20)) \leq - \frac{3}{20}

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}{20}) + 2 πn \leq \frac{π}{20} (x - 20) \leq 2 π - arccos (- \frac{3}{20}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arccos (- \frac{3}{20}) + 2 πn \leq \frac{π}{20} (x - 20) and \frac{π}{20} (x - 20) \leq 2 π - arccos (- \frac{3}{20}) + 2 πn

arccos (- \frac{3}{20}) + 2 πn \leq \frac{π}{20} (x - 20) : x \geq \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n

arccos (- \frac{3}{20}) + 2 πn \leq \frac{π}{20} (x - 20)

交换两边

\frac{π}{20} (x - 20) \geq arccos (- \frac{3}{20}) + 2 πn

在两边乘以

20

\frac{π}{20} (x - 20) \geq arccos (- \frac{3}{20}) + 2 πn

在两边乘以

20

20 \cdot \frac{π}{20} (x - 20) \geq 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) \geq 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) : π (x - 20)

20 \cdot \frac{π}{20} (x - 20)

分式相乘:

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

= \frac{20 π}{20} (x - 20)

约分:

20

= (x - 20) π

化简

20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn : 20 arccos (- \frac{3}{20}) + 40 πn

20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

数字相乘:

20 \cdot 2 = 40

= 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \geq 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \geq 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \geq 20 arccos (- \frac{3}{20}) + 40 πn

两边除以

π

π (x - 20) \geq 20 arccos (- \frac{3}{20}) + 40 πn

两边除以

π

\frac{π ( x - 20 )}{π} \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π}

化简

x - 20 \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

x - 20 \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

将

20

到右边

x - 20 \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

两边加上

20

x - 20 + 20 \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

化简

x \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

x \geq \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

化简

\frac{20 arccos ( - \frac{3}{20} )}{π} + 20 : \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π}

\frac{20 arccos ( - \frac{3}{20} )}{π} + 20

将项转换为分式:

20 = \frac{20 π}{π}

= \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{20 π}{π}

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{20 arccos ( - \frac{3}{20} ) + 20 π}{π}

x \geq \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n

\frac{π}{20} (x - 20) \leq 2 π - arccos (- \frac{3}{20}) + 2 πn : x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

\frac{π}{20} (x - 20) \leq 2 π - arccos (- \frac{3}{20}) + 2 πn

在两边乘以

20

\frac{π}{20} (x - 20) \leq 2 π - arccos (- \frac{3}{20}) + 2 πn

在两边乘以

20

20 \cdot \frac{π}{20} (x - 20) \leq 20 \cdot 2 π - 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) \leq 20 \cdot 2 π - 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

化简

20 \cdot \frac{π}{20} (x - 20) : π (x - 20)

20 \cdot \frac{π}{20} (x - 20)

分式相乘:

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

= \frac{20 π}{20} (x - 20)

约分:

20

= (x - 20) π

化简

20 \cdot 2 π - 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn : 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

20 \cdot 2 π - 20 arccos (- \frac{3}{20}) + 20 \cdot 2 πn

数字相乘:

20 \cdot 2 = 40

= 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \leq 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \leq 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

π (x - 20) \leq 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

两边除以

π

π (x - 20) \leq 40 π - 20 arccos (- \frac{3}{20}) + 40 πn

两边除以

π

\frac{π ( x - 20 )}{π} \leq \frac{40 π}{π} - \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π}

化简

\frac{π ( x - 20 )}{π} \leq \frac{40 π}{π} - \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π}

化简

\frac{π ( x - 20 )}{π} : x - 20

\frac{π ( x - 20 )}{π}

约分:

π

= x - 20

化简

\frac{40 π}{π} - \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π} : 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

\frac{40 π}{π} - \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π}

消掉

\frac{40 π}{π} : 40

\frac{40 π}{π}

约分:

π

= 40

= 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + \frac{40 πn}{π}

消掉

\frac{40 πn}{π} : 40 n

\frac{40 πn}{π}

约分:

π

= 40 n

= 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

x - 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

x - 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

x - 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

将

20

到右边

x - 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n

两边加上

20

x - 20 + 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

化简

x - 20 + 20 \leq 40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

化简

x - 20 + 20 : x

x - 20 + 20

同类项相加:

- 20 + 20 \leq 0

= x

化简

40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20 : 40 n + 60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

40 - \frac{20 arccos ( - \frac{3}{20} )}{π} + 40 n + 20

数字相加:

40 + 20 = 60

= 40 n + 60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

x \leq 40 n + 60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

x \leq 40 n + 60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

x \leq 40 n + 60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

化简

60 - \frac{20 arccos ( - \frac{3}{20} )}{π} : \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π}

60 - \frac{20 arccos ( - \frac{3}{20} )}{π}

将项转换为分式:

60 = \frac{60 π}{π}

= \frac{60 π}{π} - \frac{20 arccos ( - \frac{3}{20} )}{π}

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{60 π - 20 arccos ( - \frac{3}{20} )}{π}

x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

合并区间

x \geq \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n and x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

合并重叠的区间

\frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n \leq x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

合并区间

\frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n and \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n \leq x \leq \frac{60 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n

合并重叠的区间

\frac{- 20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 π - 20 arccos ( - \frac{3}{20} )}{π} + 40 n or \frac{20 arccos ( - \frac{3}{20} ) + 20 π}{π} + 40 n \leq x \leq \frac{20 arccos ( - \frac{9}{10} ) + 20 π}{π} + 40 n

流行的例子

tan(θ)=-1\land sin(θ)>0

tan (θ) = - 1 and sin (θ) > 0

sin(x/2-pi/3)0<= x<= 2pi

sin (\frac{x}{2} - \frac{π}{3}) 0 \leq x \leq 2 π

cos(θ)= 1/4 \land 0>θ>90,tan(θ)

cos (θ) = \frac{1}{4} and 0^{\circ} > θ > 9 0^{\circ}, tan (θ)

csc(θ)>0\land cot(θ)>0

csc (θ) > 0 and cot (θ) > 0

-(sqrt(2))/2 <sin(x)<(sqrt(2))/2

- \frac{2}{2} < sin (x) < \frac{2}{2}