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受欢迎的三角函数 >

7cos(9x)-1=1

解答

7 cos (9 x) - 1 = 1

解答

x = \frac{1.28104 \dots}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{1.28104 \dots}{9} + \frac{2 πn}{9}

+1

度数

x = 8.15538 \dots^{\circ} + 4 0^{\circ} n, x = 31.84461 \dots^{\circ} + 4 0^{\circ} n

求解步骤

7 cos (9 x) - 1 = 1

将

1

到右边

7 cos (9 x) - 1 = 1

两边加上

1

7 cos (9 x) - 1 + 1 = 1 + 1

化简

7 cos (9 x) = 2

7 cos (9 x) = 2

两边除以

7

7 cos (9 x) = 2

两边除以

7

\frac{7 cos ( 9 x )}{7} = \frac{2}{7}

化简

cos (9 x) = \frac{2}{7}

cos (9 x) = \frac{2}{7}

使用反三角函数性质

cos (9 x) = \frac{2}{7}

cos (9 x) = \frac{2}{7}

的通解

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

9 x = arccos (\frac{2}{7}) + 2 πn, 9 x = 2 π - arccos (\frac{2}{7}) + 2 πn

9 x = arccos (\frac{2}{7}) + 2 πn, 9 x = 2 π - arccos (\frac{2}{7}) + 2 πn

解

9 x = arccos (\frac{2}{7}) + 2 πn : x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

9 x = arccos (\frac{2}{7}) + 2 πn

两边除以

9

9 x = arccos (\frac{2}{7}) + 2 πn

两边除以

9

\frac{9 x}{9} = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

化简

x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

解

9 x = 2 π - arccos (\frac{2}{7}) + 2 πn : x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

9 x = 2 π - arccos (\frac{2}{7}) + 2 πn

两边除以

9

9 x = 2 π - arccos (\frac{2}{7}) + 2 πn

两边除以

9

\frac{9 x}{9} = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

化简

x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

以小数形式表示解

x = \frac{1.28104 \dots}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{1.28104 \dots}{9} + \frac{2 πn}{9}

作图

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