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受欢迎的三角函数 >

solvefor t,0.08=0.1cos(4t)

解答

求解

t, 0.08 = 0.1 cos (4 t)

解答

t = \frac{0.64350 \dots}{4} + \frac{πn}{2}, t = \frac{π}{2} - \frac{0.64350 \dots}{4} + \frac{πn}{2}

+1

度数

t = 9.21747 \dots^{\circ} + 9 0^{\circ} n, t = 80.78252 \dots^{\circ} + 9 0^{\circ} n

求解步骤

0.08 = 0.1 cos (4 t)

交换两边

0.1 cos (4 t) = 0.08

在两边乘以

100

0.1 cos (4 t) = 0.08

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

0.1 cos (4 t) \cdot 100 = 0.08 \cdot 100

整理后得

10 cos (4 t) = 8

10 cos (4 t) = 8

两边除以

10

10 cos (4 t) = 8

两边除以

10

\frac{10 cos ( 4 t )}{10} = \frac{8}{10}

化简

cos (4 t) = \frac{4}{5}

cos (4 t) = \frac{4}{5}

使用反三角函数性质

cos (4 t) = \frac{4}{5}

cos (4 t) = \frac{4}{5}

的通解

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

4 t = arccos (\frac{4}{5}) + 2 πn, 4 t = 2 π - arccos (\frac{4}{5}) + 2 πn

4 t = arccos (\frac{4}{5}) + 2 πn, 4 t = 2 π - arccos (\frac{4}{5}) + 2 πn

解

4 t = arccos (\frac{4}{5}) + 2 πn : t = \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

4 t = arccos (\frac{4}{5}) + 2 πn

两边除以

4

4 t = arccos (\frac{4}{5}) + 2 πn

两边除以

4

\frac{4 t}{4} = \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4}

化简

t = \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

t = \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

解

4 t = 2 π - arccos (\frac{4}{5}) + 2 πn : t = \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

4 t = 2 π - arccos (\frac{4}{5}) + 2 πn

两边除以

4

4 t = 2 π - arccos (\frac{4}{5}) + 2 πn

两边除以

4

\frac{4 t}{4} = \frac{2 π}{4} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4}

化简

\frac{4 t}{4} = \frac{2 π}{4} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4}

化简

\frac{4 t}{4} : t

\frac{4 t}{4}

数字相除:

\frac{4}{4} = 1

= t

化简

\frac{2 π}{4} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4} : \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

\frac{2 π}{4} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4}

消掉

\frac{2 π}{4} : \frac{π}{2}

\frac{2 π}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{2 πn}{4}

消掉

\frac{2 πn}{4} : \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

t = \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

t = \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

t = \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

t = \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}, t = \frac{π}{2} - \frac{arccos ( \frac{4}{5} )}{4} + \frac{πn}{2}

以小数形式表示解

t = \frac{0.64350 \dots}{4} + \frac{πn}{2}, t = \frac{π}{2} - \frac{0.64350 \dots}{4} + \frac{πn}{2}

作图

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