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受欢迎的三角函数 >

13<2.5cos(((2pi)/(365))(x+9))+12

解答

13 < 2.5 cos ((\frac{2 π}{365}) (x + 9)) + 12

解答

\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n < x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

+2

间隔符号

(\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n, \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n)

十进制

- 76.34434 \dots + 365 n < x < 58.34434 \dots + 365 n

求解步骤

13 < 2.5 cos (\frac{2 π}{365} (x + 9)) + 12

交换两边

2.5 cos (\frac{2 π}{365} (x + 9)) + 12 > 13

在两边乘以

10

2.5 cos (\frac{2 π}{365} (x + 9)) + 12 > 13

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

2.5 cos (\frac{2 π}{365} (x + 9)) \cdot 10 + 12 \cdot 10 > 13 \cdot 10

整理后得

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

将

120

到右边

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

两边减去

120

25 cos (\frac{2 π}{365} (x + 9)) + 120 - 120 > 130 - 120

化简

25 cos (\frac{2 π}{365} (x + 9)) > 10

25 cos (\frac{2 π}{365} (x + 9)) > 10

两边除以

25

25 cos (\frac{2 π}{365} (x + 9)) > 10

两边除以

25

\frac{25 cos ( \frac{2 π}{365} ( x + 9 ) )}{25} > \frac{10}{25}

化简

cos (\frac{2 π}{365} (x + 9)) > \frac{2}{5}

cos (\frac{2 π}{365} (x + 9)) > \frac{2}{5}

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

若

a < u < b

，则

a < u and u < b

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) and \frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) : x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9)

交换两边

\frac{2 π}{365} (x + 9) > - arccos (\frac{2}{5}) + 2 πn

在两边乘以

365

\frac{2 π}{365} (x + 9) > - arccos (\frac{2}{5}) + 2 πn

在两边乘以

365

365 \cdot \frac{2 π}{365} (x + 9) > - 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

化简

365 \cdot \frac{2 π}{365} (x + 9) > - 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

化简

365 \cdot \frac{2 π}{365} (x + 9) : 2 π (x + 9)

365 \cdot \frac{2 π}{365} (x + 9)

分式相乘:

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

= \frac{2 \cdot 365 π}{365} (x + 9)

约分:

365

= (x + 9) \cdot 2 π

化简

- 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn : - 365 arccos (\frac{2}{5}) + 730 πn

- 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

数字相乘:

365 \cdot 2 = 730

= - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

两边除以

2 π

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

两边除以

2 π

\frac{2 π ( x + 9 )}{2 π} > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

化简

\frac{2 π ( x + 9 )}{2 π} > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

化简

\frac{2 π ( x + 9 )}{2 π} : x + 9

\frac{2 π ( x + 9 )}{2 π}

数字相除:

\frac{2}{2} = 1

= \frac{π ( x + 9 )}{π}

约分:

π

= x + 9

化简

- \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π} : - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

- \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

消掉

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

消掉

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

数字相除:

\frac{730}{2} = 365

= \frac{365 πn}{π}

约分:

π

= 365 n

= 365 n

= - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

将

9

到右边

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

两边减去

9

x + 9 - 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

化简

x > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

x > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

化简

- \frac{365 arccos ( \frac{2}{5} )}{2 π} - 9 : \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

- \frac{365 arccos ( \frac{2}{5} )}{2 π} - 9

将项转换为分式:

9 = \frac{9 \cdot 2 π}{2 π}

= - \frac{365 arccos ( \frac{2}{5} )}{2 π} - \frac{9 \cdot 2 π}{2 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{- 365 arccos ( \frac{2}{5} ) - 9 \cdot 2 π}{2 π}

数字相乘:

9 \cdot 2 = 18

= \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn : x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

在两边乘以

365

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

在两边乘以

365

365 \cdot \frac{2 π}{365} (x + 9) < 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

化简

365 \cdot \frac{2 π}{365} (x + 9) < 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

化简

365 \cdot \frac{2 π}{365} (x + 9) : 2 π (x + 9)

365 \cdot \frac{2 π}{365} (x + 9)

分式相乘:

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

= \frac{2 \cdot 365 π}{365} (x + 9)

约分:

365

= (x + 9) \cdot 2 π

化简

365 arccos (\frac{2}{5}) + 365 \cdot 2 πn : 365 arccos (\frac{2}{5}) + 730 πn

365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

数字相乘:

365 \cdot 2 = 730

= 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

两边除以

2 π

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

两边除以

2 π

\frac{2 π ( x + 9 )}{2 π} < \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

化简

\frac{2 π ( x + 9 )}{2 π} < \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

化简

\frac{2 π ( x + 9 )}{2 π} : x + 9

\frac{2 π ( x + 9 )}{2 π}

数字相除:

\frac{2}{2} = 1

= \frac{π ( x + 9 )}{π}

约分:

π

= x + 9

化简

\frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π} : \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

\frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

消掉

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

消掉

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

数字相除:

\frac{730}{2} = 365

= \frac{365 πn}{π}

约分:

π

= 365 n

= 365 n

= \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

将

9

到右边

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

两边减去

9

x + 9 - 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

化简

x < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

x < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

化简

\frac{365 arccos ( \frac{2}{5} )}{2 π} - 9 : \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

\frac{365 arccos ( \frac{2}{5} )}{2 π} - 9

将项转换为分式:

9 = \frac{9 \cdot 2 π}{2 π}

= \frac{365 arccos ( \frac{2}{5} )}{2 π} - \frac{9 \cdot 2 π}{2 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{365 arccos ( \frac{2}{5} ) - 9 \cdot 2 π}{2 π}

数字相乘:

9 \cdot 2 = 18

= \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

合并区间

x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n and x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

合并重叠的区间

\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n < x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

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