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受欢迎的三角函数 >

cos(4x)+sin(x/2)=0

解答

cos (4 x) + sin (\frac{x}{2}) = 0

解答

x \in R 无解

求解步骤

cos (4 x) + sin (\frac{x}{2}) = 0

两边减去

sin (\frac{x}{2})

cos (4 x) = - sin (\frac{x}{2})

使用三角恒等式改写

cos (4 x) = - sin (\frac{x}{2})

利用以下特性:

- sin (x) = sin (- x)

cos (4 x) = sin (- (\frac{x}{2}))

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (\frac{π}{2} - 4 x) = sin (- (\frac{x}{2}))

sin (\frac{π}{2} - 4 x) = sin (- (\frac{x}{2}))

使用反三角函数性质

sin (\frac{π}{2} - 4 x) = sin (- (\frac{x}{2}))

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

- (\frac{x}{2}) = \frac{π}{2} - 4 x + 2 πn, - (\frac{x}{2}) = π - (\frac{π}{2} - 4 x) + 2 πn

- (\frac{x}{2}) = \frac{π}{2} - 4 x + 2 πn, - (\frac{x}{2}) = π - (\frac{π}{2} - 4 x) + 2 πn

- (\frac{x}{2}) = \frac{π}{2} - 4 x + 2 πn : x = \frac{π + 4 πn}{7}

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

将

4 x

para o lado esquerdo

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

两边加上

4 x

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

化简

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

化简

- \frac{x}{2} + 4 x : \frac{7 x}{2}

- \frac{x}{2} + 4 x

将项转换为分式:

4 x = \frac{4 x 2}{2}

= - \frac{x}{2} + \frac{4 x \cdot 2}{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{- x + 4 x \cdot 2}{2}

- x + 4 x \cdot 2 = 7 x

- x + 4 x \cdot 2

数字相乘:

4 \cdot 2 = 8

= - x + 8 x

同类项相加:

- x + 8 x = 7 x

= 7 x

= \frac{7 x}{2}

化简

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

\frac{π}{2} - 4 x + 2 πn + 4 x

同类项相加:

- 4 x + 4 x = 0

= \frac{π}{2} + 2 πn

\frac{7 x}{2} = \frac{π}{2} + 2 πn

\frac{7 x}{2} = \frac{π}{2} + 2 πn

\frac{7 x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{7 x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 \cdot 7 x}{2} : 7 x

\frac{2 \cdot 7 x}{2}

数字相乘:

2 \cdot 7 = 14

= \frac{14 x}{2}

数字相除:

\frac{14}{2} = 7

= 7 x

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

7 x = π + 4 πn

7 x = π + 4 πn

7 x = π + 4 πn

两边除以

7

7 x = π + 4 πn

两边除以

7

\frac{7 x}{7} = \frac{π}{7} + \frac{4 πn}{7}

化简

\frac{7 x}{7} = \frac{π}{7} + \frac{4 πn}{7}

化简

\frac{7 x}{7} : x

\frac{7 x}{7}

数字相除:

\frac{7}{7} = 1

= x

化简

\frac{π}{7} + \frac{4 πn}{7} : \frac{π + 4 πn}{7}

\frac{π}{7} + \frac{4 πn}{7}

使用法则

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

= \frac{π + 4 πn}{7}

x = \frac{π + 4 πn}{7}

x = \frac{π + 4 πn}{7}

x = \frac{π + 4 πn}{7}

- (\frac{x}{2}) = π - (\frac{π}{2} - 4 x) + 2 πn : x = - \frac{π + 4 πn}{9}

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

将

(\frac{π}{2} - 4 x)

para o lado esquerdo

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

两边加上

(\frac{π}{2} - 4 x)

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

化简

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

化简

- \frac{x}{2} + \frac{π}{2} - 4 x : \frac{- 9 x + π}{2}

- \frac{x}{2} + \frac{π}{2} - 4 x

合并分式

- \frac{x}{2} + \frac{π}{2} : \frac{- x + π}{2}

使用法则

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

= \frac{- x + π}{2}

= \frac{- x + π}{2} - 4 x

将项转换为分式:

4 x = \frac{4 x 2}{2}

= \frac{- x + π}{2} - \frac{4 x \cdot 2}{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{- x + π - 4 x \cdot 2}{2}

- x + π - 4 x \cdot 2 = - 9 x + π

- x + π - 4 x \cdot 2

数字相乘:

4 \cdot 2 = 8

= - x + π - 8 x

对同类项分组

= - x - 8 x + π

同类项相加:

- x - 8 x = - 9 x

= - 9 x + π

= \frac{- 9 x + π}{2}

化简

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

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

同类项相加:

- (\frac{π}{2} - 4 x) + \frac{π}{2} - 4 x = 0

= π + 2 πn

\frac{- 9 x + π}{2} = π + 2 πn

\frac{- 9 x + π}{2} = π + 2 πn

\frac{- 9 x + π}{2} = π + 2 πn

在两边乘以

2

\frac{- 9 x + π}{2} = π + 2 πn

在两边乘以

2

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

化简

- 9 x + π = 2 π + 4 πn

- 9 x + π = 2 π + 4 πn

将

π

到右边

- 9 x + π = 2 π + 4 πn

两边减去

π

- 9 x + π - π = 2 π + 4 πn - π

化简

- 9 x = π + 4 πn

- 9 x = π + 4 πn

两边除以

- 9

- 9 x = π + 4 πn

两边除以

- 9

\frac{- 9 x}{- 9} = \frac{π}{- 9} + \frac{4 πn}{- 9}

化简

\frac{- 9 x}{- 9} = \frac{π}{- 9} + \frac{4 πn}{- 9}

化简

\frac{- 9 x}{- 9} : x

\frac{- 9 x}{- 9}

使用分式法则:

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

= \frac{9 x}{9}

数字相除:

\frac{9}{9} = 1

= x

化简

\frac{π}{- 9} + \frac{4 πn}{- 9} : - \frac{π + 4 πn}{9}

\frac{π}{- 9} + \frac{4 πn}{- 9}

使用法则

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

= \frac{π + 4 πn}{- 9}

使用分式法则:

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

= - \frac{π + 4 πn}{9}

x = - \frac{π + 4 πn}{9}

x = - \frac{π + 4 πn}{9}

x = - \frac{π + 4 πn}{9}

因为方程对以下值无定义:

\frac{π + 4 πn}{7}, - \frac{π + 4 πn}{9}

x \in R 无解

作图

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