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受欢迎的三角函数 >

solvefor t,F=4cos(-2t+7)+C

解答

求解

t, F = 4 cos (- 2 t + 7) + C

解答

t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}, t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

求解步骤

F = 4 cos (- 2 t + 7) + C

交换两边

4 cos (- 2 t + 7) + C = F

将

C

到右边

4 cos (- 2 t + 7) + C = F

两边减去

C

4 cos (- 2 t + 7) + C - C = F - C

化简

4 cos (- 2 t + 7) = F - C

4 cos (- 2 t + 7) = F - C

两边除以

4

4 cos (- 2 t + 7) = F - C

两边除以

4

\frac{4 cos ( - 2 t + 7 )}{4} = \frac{F}{4} - \frac{C}{4}

化简

\frac{4 cos ( - 2 t + 7 )}{4} = \frac{F}{4} - \frac{C}{4}

化简

\frac{4 cos ( - 2 t + 7 )}{4} : cos (- 2 t + 7)

\frac{4 cos ( - 2 t + 7 )}{4}

数字相除:

\frac{4}{4} = 1

= cos (- 2 t + 7)

化简

\frac{F}{4} - \frac{C}{4} : \frac{F - C}{4}

\frac{F}{4} - \frac{C}{4}

使用法则

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

= \frac{F - C}{4}

cos (- 2 t + 7) = \frac{F - C}{4}

cos (- 2 t + 7) = \frac{F - C}{4}

cos (- 2 t + 7) = \frac{F - C}{4}

使用反三角函数性质

cos (- 2 t + 7) = \frac{F - C}{4}

cos (- 2 t + 7) = \frac{F - C}{4}

的通解

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

- 2 t + 7 = arccos (\frac{F - C}{4}) + 2 πn, - 2 t + 7 = - arccos (\frac{F - C}{4}) + 2 πn

- 2 t + 7 = arccos (\frac{F - C}{4}) + 2 πn, - 2 t + 7 = - arccos (\frac{F - C}{4}) + 2 πn

解

- 2 t + 7 = arccos (\frac{F - C}{4}) + 2 πn : t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

- 2 t + 7 = arccos (\frac{F - C}{4}) + 2 πn

将

7

到右边

- 2 t + 7 = arccos (\frac{F - C}{4}) + 2 πn

两边减去

7

- 2 t + 7 - 7 = arccos (\frac{F - C}{4}) + 2 πn - 7

化简

- 2 t = arccos (\frac{F - C}{4}) + 2 πn - 7

- 2 t = arccos (\frac{F - C}{4}) + 2 πn - 7

两边除以

- 2

- 2 t = arccos (\frac{F - C}{4}) + 2 πn - 7

两边除以

- 2

\frac{- 2 t}{- 2} = \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

化简

\frac{- 2 t}{- 2} = \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

化简

\frac{- 2 t}{- 2} : t

\frac{- 2 t}{- 2}

使用分式法则:

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

= \frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

\frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2} : - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

\frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

使用分式法则:

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

= - \frac{arccos ( \frac{F - C}{4} )}{2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

使用分式法则:

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

= - \frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= - πn

= - \frac{arccos ( \frac{F - C}{4} )}{2} - πn - \frac{7}{- 2}

使用分式法则:

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

= - \frac{arccos ( \frac{F - C}{4} )}{2} - πn - (- \frac{7}{2})

使用法则

- (- a) = a

= - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

解

- 2 t + 7 = - arccos (\frac{F - C}{4}) + 2 πn : t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

- 2 t + 7 = - arccos (\frac{F - C}{4}) + 2 πn

将

7

到右边

- 2 t + 7 = - arccos (\frac{F - C}{4}) + 2 πn

两边减去

7

- 2 t + 7 - 7 = - arccos (\frac{F - C}{4}) + 2 πn - 7

化简

- 2 t = - arccos (\frac{F - C}{4}) + 2 πn - 7

- 2 t = - arccos (\frac{F - C}{4}) + 2 πn - 7

两边除以

- 2

- 2 t = - arccos (\frac{F - C}{4}) + 2 πn - 7

两边除以

- 2

\frac{- 2 t}{- 2} = - \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

化简

\frac{- 2 t}{- 2} = - \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

化简

\frac{- 2 t}{- 2} : t

\frac{- 2 t}{- 2}

使用分式法则:

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

= \frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

- \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2} : \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

- \frac{arccos ( \frac{F - C}{4} )}{- 2} + \frac{2 πn}{- 2} - \frac{7}{- 2}

\frac{arccos ( \frac{F - C}{4} )}{- 2} = - \frac{arccos ( \frac{F - C}{4} )}{2}

\frac{arccos ( \frac{F - C}{4} )}{- 2}

使用分式法则:

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

= - \frac{arccos ( \frac{F - C}{4} )}{2}

\frac{2 πn}{- 2} = - πn

\frac{2 πn}{- 2}

使用分式法则:

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

= - \frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= - πn

= - (- \frac{arccos ( \frac{F - C}{4} )}{2}) - πn - \frac{7}{- 2}

使用法则

- (- a) = a

= \frac{arccos ( \frac{F - C}{4} )}{2} - πn - \frac{7}{- 2}

使用分式法则:

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

= \frac{arccos ( \frac{F - C}{4} )}{2} - πn - (- \frac{7}{2})

使用法则

- (- a) = a

= \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

t = - \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}, t = \frac{arccos ( \frac{F - C}{4} )}{2} - πn + \frac{7}{2}

作图

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