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受欢迎的三角函数 >

solvefor x,80=75-60cos(([pi*x])/(15))

解答

求解

x, 80 = 75 - 60 cos (\frac{[ π \cdot x ]}{15})

解答

x = \frac{15 \cdot 1.65422 \dots}{π} + 30 n, x = - \frac{15 \cdot 1.65422 \dots}{π} + 30 n

+1

度数

x = 452.54208 \dots^{\circ} + 1718.87338 \dots^{\circ} n, x = - 452.54208 \dots^{\circ} + 1718.87338 \dots^{\circ} n

求解步骤

80 = 75 - 60 cos (\frac{[ π x ]}{15})

交换两边

75 - 60 cos (\frac{[ π x ]}{15}) = 80

将

75

到右边

75 - 60 cos (\frac{[ π x ]}{15}) = 80

两边减去

75

75 - 60 cos (\frac{[ π x ]}{15}) - 75 = 80 - 75

化简

- 60 cos (\frac{[ π x ]}{15}) = 5

- 60 cos (\frac{[ π x ]}{15}) = 5

两边除以

- 60

- 60 cos (\frac{[ π x ]}{15}) = 5

两边除以

- 60

\frac{- 60 cos ( \frac{[ π x ]}{15} )}{- 60} = \frac{5}{- 60}

化简

\frac{- 60 cos ( \frac{[ π x ]}{15} )}{- 60} = \frac{5}{- 60}

化简

\frac{- 60 cos ( \frac{[ π x ]}{15} )}{- 60} : cos (\frac{[ π x ]}{15})

\frac{- 60 cos ( \frac{[ π x ]}{15} )}{- 60}

使用分式法则:

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

= \frac{60 cos ( \frac{[ π x ]}{15} )}{60}

数字相除:

\frac{60}{60} = 1

= cos (\frac{[ π x ]}{15})

化简

\frac{5}{- 60} : - \frac{1}{12}

\frac{5}{- 60}

使用分式法则:

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

= - \frac{5}{60}

约分:

5

= - \frac{1}{12}

cos (\frac{[ π x ]}{15}) = - \frac{1}{12}

cos (\frac{[ π x ]}{15}) = - \frac{1}{12}

cos (\frac{[ π x ]}{15}) = - \frac{1}{12}

使用反三角函数性质

cos (\frac{[ π x ]}{15}) = - \frac{1}{12}

cos (\frac{[ π x ]}{15}) = - \frac{1}{12}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

\frac{[ π x ]}{15} = arccos (- \frac{1}{12}) + 2 πn, \frac{[ π x ]}{15} = - arccos (- \frac{1}{12}) + 2 πn

\frac{[ π x ]}{15} = arccos (- \frac{1}{12}) + 2 πn, \frac{[ π x ]}{15} = - arccos (- \frac{1}{12}) + 2 πn

解

\frac{[ π x ]}{15} = arccos (- \frac{1}{12}) + 2 πn : x = \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

\frac{[ π x ]}{15} = arccos (- \frac{1}{12}) + 2 πn

在两边乘以

15

\frac{[ π x ]}{15} = arccos (- \frac{1}{12}) + 2 πn

在两边乘以

15

\frac{15 [ π x ]}{15} = 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

化简

\frac{15 [ π x ]}{15} = 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

化简

\frac{15 ( π x )}{15} : π x

\frac{15 [ π x ]}{15}

去除括号:

(a) = a

= \frac{15 π x}{15}

数字相除:

\frac{15}{15} = 1

= π x

化简

15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn : 15 arccos (- \frac{1}{12}) + 30 πn

15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

数字相乘:

15 \cdot 2 = 30

= 15 arccos (- \frac{1}{12}) + 30 πn

π x = 15 arccos (- \frac{1}{12}) + 30 πn

π x = 15 arccos (- \frac{1}{12}) + 30 πn

π x = 15 arccos (- \frac{1}{12}) + 30 πn

两边除以

π

π x = 15 arccos (- \frac{1}{12}) + 30 πn

两边除以

π

\frac{π x}{π} = \frac{15 arccos ( - \frac{1}{12} )}{π} + \frac{30 πn}{π}

化简

x = \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

x = \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

解

\frac{[ π x ]}{15} = - arccos (- \frac{1}{12}) + 2 πn : x = - \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

\frac{[ π x ]}{15} = - arccos (- \frac{1}{12}) + 2 πn

在两边乘以

15

\frac{[ π x ]}{15} = - arccos (- \frac{1}{12}) + 2 πn

在两边乘以

15

\frac{15 [ π x ]}{15} = - 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

化简

\frac{15 [ π x ]}{15} = - 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

化简

\frac{15 ( π x )}{15} : π x

\frac{15 [ π x ]}{15}

去除括号:

(a) = a

= \frac{15 π x}{15}

数字相除:

\frac{15}{15} = 1

= π x

化简

- 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn : - 15 arccos (- \frac{1}{12}) + 30 πn

- 15 arccos (- \frac{1}{12}) + 15 \cdot 2 πn

数字相乘:

15 \cdot 2 = 30

= - 15 arccos (- \frac{1}{12}) + 30 πn

π x = - 15 arccos (- \frac{1}{12}) + 30 πn

π x = - 15 arccos (- \frac{1}{12}) + 30 πn

π x = - 15 arccos (- \frac{1}{12}) + 30 πn

两边除以

π

π x = - 15 arccos (- \frac{1}{12}) + 30 πn

两边除以

π

\frac{π x}{π} = - \frac{15 arccos ( - \frac{1}{12} )}{π} + \frac{30 πn}{π}

化简

x = - \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

x = - \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

x = \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n, x = - \frac{15 arccos ( - \frac{1}{12} )}{π} + 30 n

以小数形式表示解

x = \frac{15 \cdot 1.65422 \dots}{π} + 30 n, x = - \frac{15 \cdot 1.65422 \dots}{π} + 30 n

作图

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