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受欢迎的三角函数 >

1=7cos(pi/3 t)

解答

1 = 7 cos (\frac{π}{3} t)

解答

t = \frac{3 \cdot 1.42744 \dots}{π} + 6 n, t = 6 - \frac{3 \cdot 1.42744 \dots}{π} + 6 n

+1

度数

t = 78.10063 \dots^{\circ} + 343.77467 \dots^{\circ} n, t = 265.67404 \dots^{\circ} + 343.77467 \dots^{\circ} n

求解步骤

1 = 7 cos (\frac{π}{3} t)

交换两边

7 cos (\frac{π}{3} t) = 1

两边除以

7

7 cos (\frac{π}{3} t) = 1

两边除以

7

\frac{7 cos ( \frac{π}{3} t )}{7} = \frac{1}{7}

化简

cos (\frac{π}{3} t) = \frac{1}{7}

cos (\frac{π}{3} t) = \frac{1}{7}

使用反三角函数性质

cos (\frac{π}{3} t) = \frac{1}{7}

cos (\frac{π}{3} t) = \frac{1}{7}

的通解

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

\frac{π}{3} t = arccos (\frac{1}{7}) + 2 πn, \frac{π}{3} t = 2 π - arccos (\frac{1}{7}) + 2 πn

\frac{π}{3} t = arccos (\frac{1}{7}) + 2 πn, \frac{π}{3} t = 2 π - arccos (\frac{1}{7}) + 2 πn

解

\frac{π}{3} t = arccos (\frac{1}{7}) + 2 πn : t = \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

\frac{π}{3} t = arccos (\frac{1}{7}) + 2 πn

在两边乘以

3

\frac{π}{3} t = arccos (\frac{1}{7}) + 2 πn

在两边乘以

3

3 \cdot \frac{π}{3} t = 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

化简

3 \cdot \frac{π}{3} t = 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

化简

3 \cdot \frac{π}{3} t : π t

3 \cdot \frac{π}{3} t

分式相乘:

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

= \frac{3 π}{3} t

约分:

3

= t π

化简

3 arccos (\frac{1}{7}) + 3 \cdot 2 πn : 3 arccos (\frac{1}{7}) + 6 πn

3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

数字相乘:

3 \cdot 2 = 6

= 3 arccos (\frac{1}{7}) + 6 πn

π t = 3 arccos (\frac{1}{7}) + 6 πn

π t = 3 arccos (\frac{1}{7}) + 6 πn

π t = 3 arccos (\frac{1}{7}) + 6 πn

两边除以

π

π t = 3 arccos (\frac{1}{7}) + 6 πn

两边除以

π

\frac{π t}{π} = \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π}

化简

t = \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

t = \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

解

\frac{π}{3} t = 2 π - arccos (\frac{1}{7}) + 2 πn : t = 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

\frac{π}{3} t = 2 π - arccos (\frac{1}{7}) + 2 πn

在两边乘以

3

\frac{π}{3} t = 2 π - arccos (\frac{1}{7}) + 2 πn

在两边乘以

3

3 \cdot \frac{π}{3} t = 3 \cdot 2 π - 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

化简

3 \cdot \frac{π}{3} t = 3 \cdot 2 π - 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

化简

3 \cdot \frac{π}{3} t : π t

3 \cdot \frac{π}{3} t

分式相乘:

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

= \frac{3 π}{3} t

约分:

3

= t π

化简

3 \cdot 2 π - 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn : 6 π - 3 arccos (\frac{1}{7}) + 6 πn

3 \cdot 2 π - 3 arccos (\frac{1}{7}) + 3 \cdot 2 πn

数字相乘:

3 \cdot 2 = 6

= 6 π - 3 arccos (\frac{1}{7}) + 6 πn

π t = 6 π - 3 arccos (\frac{1}{7}) + 6 πn

π t = 6 π - 3 arccos (\frac{1}{7}) + 6 πn

π t = 6 π - 3 arccos (\frac{1}{7}) + 6 πn

两边除以

π

π t = 6 π - 3 arccos (\frac{1}{7}) + 6 πn

两边除以

π

\frac{π t}{π} = \frac{6 π}{π} - \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π}

化简

\frac{π t}{π} = \frac{6 π}{π} - \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{6 π}{π} - \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π} : 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

\frac{6 π}{π} - \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π}

消掉

\frac{6 π}{π} : 6

\frac{6 π}{π}

约分:

π

= 6

= 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + \frac{6 πn}{π}

消掉

\frac{6 πn}{π} : 6 n

\frac{6 πn}{π}

约分:

π

= 6 n

= 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

t = 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

t = 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

t = 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

t = \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n, t = 6 - \frac{3 arccos ( \frac{1}{7} )}{π} + 6 n

以小数形式表示解

t = \frac{3 \cdot 1.42744 \dots}{π} + 6 n, t = 6 - \frac{3 \cdot 1.42744 \dots}{π} + 6 n

作图

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