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受欢迎的三角函数 >

cos(6x)=sin(x-1)

解答

cos (6 x) = sin (x - 1)

解答

x = \frac{2 + 4 πn + π}{14}, x = - \frac{π + 4 πn + 2}{10}

+1

度数

x = 21.04225 \dots^{\circ} + 51.42857 \dots^{\circ} n, x = - 29.45915 \dots^{\circ} - 7 2^{\circ} n

求解步骤

cos (6 x) = sin (x - 1)

使用三角恒等式改写

cos (6 x) = sin (x - 1)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

cos (6 x) = sin (\frac{π}{2} - 6 x)

cos (6 x) = sin (\frac{π}{2} - 6 x)

使用反三角函数性质

cos (6 x) = sin (\frac{π}{2} - 6 x)

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

x - 1 = \frac{π}{2} - 6 x + 2 πn, x - 1 = π - (\frac{π}{2} - 6 x) + 2 πn

x - 1 = \frac{π}{2} - 6 x + 2 πn, x - 1 = π - (\frac{π}{2} - 6 x) + 2 πn

x - 1 = \frac{π}{2} - 6 x + 2 πn : x = \frac{2 + 4 πn + π}{14}

x - 1 = \frac{π}{2} - 6 x + 2 πn

将

1

到右边

x - 1 = \frac{π}{2} - 6 x + 2 πn

两边加上

1

x - 1 + 1 = \frac{π}{2} - 6 x + 2 πn + 1

化简

x = \frac{π}{2} - 6 x + 2 πn + 1

x = \frac{π}{2} - 6 x + 2 πn + 1

将

6 x

para o lado esquerdo

x = \frac{π}{2} - 6 x + 2 πn + 1

两边加上

6 x

x + 6 x = \frac{π}{2} - 6 x + 2 πn + 1 + 6 x

化简

7 x = \frac{π}{2} + 2 πn + 1

7 x = \frac{π}{2} + 2 πn + 1

两边除以

7

7 x = \frac{π}{2} + 2 πn + 1

两边除以

7

\frac{7 x}{7} = \frac{\frac{π}{2}}{7} + \frac{2 πn}{7} + \frac{1}{7}

化简

\frac{7 x}{7} = \frac{\frac{π}{2}}{7} + \frac{2 πn}{7} + \frac{1}{7}

化简

\frac{7 x}{7} : x

\frac{7 x}{7}

数字相除:

\frac{7}{7} = 1

= x

化简

\frac{\frac{π}{2}}{7} + \frac{2 πn}{7} + \frac{1}{7} : \frac{2 + 4 πn + π}{14}

\frac{\frac{π}{2}}{7} + \frac{2 πn}{7} + \frac{1}{7}

对同类项分组

= \frac{1}{7} + \frac{2 πn}{7} + \frac{\frac{π}{2}}{7}

使用法则

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

= \frac{1 + 2 πn + \frac{π}{2}}{7}

化简

1 + 2 πn + \frac{π}{2} : \frac{2 + 4 πn + π}{2}

1 + 2 πn + \frac{π}{2}

将项转换为分式:

1 = \frac{1 \cdot 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{1 \cdot 2}{2} + \frac{2 πn \cdot 2}{2} + \frac{π}{2}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot 2 + 2 πn \cdot 2 + π}{2}

1 \cdot 2 + 2 πn \cdot 2 + π = 2 + 4 πn + π

1 \cdot 2 + 2 πn \cdot 2 + π

数字相乘:

1 \cdot 2 = 2

= 2 + 2 \cdot 2 πn + π

数字相乘:

2 \cdot 2 = 4

= 2 + 4 πn + π

= \frac{2 + 4 πn + π}{2}

= \frac{\frac{2 + 4 πn + π}{2}}{7}

使用分式法则:

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

= \frac{2 + 4 πn + π}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{2 + 4 πn + π}{14}

x = \frac{2 + 4 πn + π}{14}

x = \frac{2 + 4 πn + π}{14}

x = \frac{2 + 4 πn + π}{14}

x - 1 = π - (\frac{π}{2} - 6 x) + 2 πn : x = - \frac{π + 4 πn + 2}{10}

x - 1 = π - (\frac{π}{2} - 6 x) + 2 πn

展开

π - (\frac{π}{2} - 6 x) + 2 πn : π - \frac{π}{2} + 6 x + 2 πn

π - (\frac{π}{2} - 6 x) + 2 πn

- (\frac{π}{2} - 6 x) : - \frac{π}{2} + 6 x

- (\frac{π}{2} - 6 x)

打开括号

= - (\frac{π}{2}) - (- 6 x)

使用加减运算法则

- (- a) = a, - (a) = - a

= - \frac{π}{2} + 6 x

= π - \frac{π}{2} + 6 x + 2 πn

x - 1 = π - \frac{π}{2} + 6 x + 2 πn

将

1

到右边

x - 1 = π - \frac{π}{2} + 6 x + 2 πn

两边加上

1

x - 1 + 1 = π - \frac{π}{2} + 6 x + 2 πn + 1

化简

x = π - \frac{π}{2} + 6 x + 2 πn + 1

x = π - \frac{π}{2} + 6 x + 2 πn + 1

将

6 x

para o lado esquerdo

x = π - \frac{π}{2} + 6 x + 2 πn + 1

两边减去

6 x

x - 6 x = π - \frac{π}{2} + 6 x + 2 πn + 1 - 6 x

化简

- 5 x = π - \frac{π}{2} + 2 πn + 1

- 5 x = π - \frac{π}{2} + 2 πn + 1

两边除以

- 5

- 5 x = π - \frac{π}{2} + 2 πn + 1

两边除以

- 5

\frac{- 5 x}{- 5} = \frac{π}{- 5} - \frac{\frac{π}{2}}{- 5} + \frac{2 πn}{- 5} + \frac{1}{- 5}

化简

\frac{- 5 x}{- 5} = \frac{π}{- 5} - \frac{\frac{π}{2}}{- 5} + \frac{2 πn}{- 5} + \frac{1}{- 5}

化简

\frac{- 5 x}{- 5} : x

\frac{- 5 x}{- 5}

使用分式法则:

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

= \frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

\frac{π}{- 5} - \frac{\frac{π}{2}}{- 5} + \frac{2 πn}{- 5} + \frac{1}{- 5} : - \frac{π + 4 πn + 2}{10}

\frac{π}{- 5} - \frac{\frac{π}{2}}{- 5} + \frac{2 πn}{- 5} + \frac{1}{- 5}

对同类项分组

= \frac{π}{- 5} + \frac{1}{- 5} + \frac{2 πn}{- 5} - \frac{\frac{π}{2}}{- 5}

使用法则

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

= \frac{π + 1 + 2 πn - \frac{π}{2}}{- 5}

使用分式法则:

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

= - \frac{π + 1 + 2 πn - \frac{π}{2}}{5}

化简

π + 1 + 2 πn - \frac{π}{2} : \frac{π + 4 πn + 2}{2}

π + 1 + 2 πn - \frac{π}{2}

将项转换为分式:

π = \frac{π 2}{2}, 1 = \frac{1 \cdot 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{π 2}{2} + \frac{1 \cdot 2}{2} + \frac{2 πn \cdot 2}{2} - \frac{π}{2}

因为分母相等，所以合并分式:

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

= \frac{π 2 + 1 \cdot 2 + 2 πn \cdot 2 - π}{2}

π 2 + 1 \cdot 2 + 2 πn \cdot 2 - π = π + 4 πn + 2

π 2 + 1 \cdot 2 + 2 πn \cdot 2 - π

对同类项分组

= 2 π - π + 2 \cdot 2 πn + 1 \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn + 1 \cdot 2

数字相乘:

2 \cdot 2 = 4

= π + 4 πn + 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= π + 4 πn + 2

= \frac{π + 4 πn + 2}{2}

= - \frac{\frac{π + 4 πn + 2}{2}}{5}

化简

\frac{\frac{π + 4 πn + 2}{2}}{5} : \frac{π + 4 πn + 2}{10}

\frac{\frac{π + 4 πn + 2}{2}}{5}

使用分式法则:

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

= \frac{π + 4 πn + 2}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{π + 4 πn + 2}{10}

= - \frac{π + 4 πn + 2}{10}

x = - \frac{π + 4 πn + 2}{10}

x = - \frac{π + 4 πn + 2}{10}

x = - \frac{π + 4 πn + 2}{10}

x = \frac{2 + 4 πn + π}{14}, x = - \frac{π + 4 πn + 2}{10}

x = \frac{2 + 4 πn + π}{14}, x = - \frac{π + 4 πn + 2}{10}

作图

流行的例子

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