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

+2

间隔符号

(\frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n, \frac{712 π - 365 arccos ( \frac{2}{5} )}{2 π} + 365 n)

十进制

58.34434 \dots + 365 n < x < 288.65565 \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 < a \leq 1

，则

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

arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) < 2 π - 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) < 2 π - 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) < 2 π - arccos (\frac{2}{5}) + 2 πn : x < \frac{712 π - 365 arccos ( \frac{2}{5} )}{2 π} + 365 n

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

在两边乘以

365

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

在两边乘以

365

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

化简

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

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

数字相乘:

365 \cdot 2 = 730

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

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

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

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

两边除以

2 π

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

两边除以

2 π

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

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

消掉

\frac{730 π}{2 π} : 365

\frac{730 π}{2 π}

消掉

\frac{730 π}{2 π} : 365

\frac{730 π}{2 π}

数字相除:

\frac{730}{2} = 365

= \frac{365 π}{π}

约分:

π

= 365

= 365

= 365 - \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

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

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

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

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

将

9

到右边

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

两边减去

9

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

化简

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

化简

x + 9 - 9 : x

x + 9 - 9

同类项相加:

9 - 9 < 0

= x

化简

365 - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9 : 365 n + 356 - \frac{365 arccos ( \frac{2}{5} )}{2 π}

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

数字相减:

365 - 9 = 356

= 365 n + 356 - \frac{365 arccos ( \frac{2}{5} )}{2 π}

x < 365 n + 356 - \frac{365 arccos ( \frac{2}{5} )}{2 π}

x < 365 n + 356 - \frac{365 arccos ( \frac{2}{5} )}{2 π}

x < 365 n + 356 - \frac{365 arccos ( \frac{2}{5} )}{2 π}

化简

356 - \frac{365 arccos ( \frac{2}{5} )}{2 π} : \frac{712 π - 365 arccos ( \frac{2}{5} )}{2 π}

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

将项转换为分式:

356 = \frac{356 \cdot 2 π}{2 π}

= \frac{356 \cdot 2 π}{2 π} - \frac{365 arccos ( \frac{2}{5} )}{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{356 \cdot 2 π - 365 arccos ( \frac{2}{5} )}{2 π}

数字相乘:

356 \cdot 2 = 712

= \frac{712 π - 365 arccos ( \frac{2}{5} )}{2 π}

x < \frac{712 π - 365 arccos ( \frac{2}{5} )}{2 π} + 365 n

合并区间

x > \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n and x < \frac{712 π - 365 arccos ( \frac{2}{5} )}{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{712 π - 365 arccos ( \frac{2}{5} )}{2 π} + 365 n

流行的例子

(9cos(t)-1/2)/4 >= 4

\frac{9 cos ( t ) - \frac{1}{2}}{4} \geq 4

solvefor N, pi/2-arctan(N)<0.0001

so l v e f or N, \frac{π}{2} - arctan (N) < 0.0001

cos (\frac{π}{6} t) < 0

cos(x)>= 0,-pi<= x<= pi

cos (x) \geq 0, - π \leq x \leq π

3cos(2x)+2< 5/2

3 cos (2 x) + 2 < \frac{5}{2}