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受欢迎的三角函数 >

sin(2x+7)=cos(4x-7)

解答

sin (2 x + 7) = cos (4 x - 7)

解答

x = \frac{4 πn + π}{12}, x = - \frac{π + 4 πn - 28}{4}

+1

度数

x = 1 5^{\circ} + 6 0^{\circ} n, x = 356.07045 \dots^{\circ} - 18 0^{\circ} n

求解步骤

sin (2 x + 7) = cos (4 x - 7)

使用三角恒等式改写

sin (2 x + 7) = cos (4 x - 7)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

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

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

使用反三角函数性质

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

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

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

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

2 x + 7 = \frac{π}{2} - (4 x - 7) + 2 πn : x = \frac{4 πn + π}{12}

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

展开

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

\frac{π}{2} - (4 x - 7) + 2 πn

- (4 x - 7) : - 4 x + 7

- (4 x - 7)

打开括号

= - (4 x) - (- 7)

使用加减运算法则

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

= - 4 x + 7

= \frac{π}{2} - 4 x + 7 + 2 πn

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

将

7

到右边

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

两边减去

7

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

化简

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

化简

2 x + 7 - 7 : 2 x

2 x + 7 - 7

同类项相加:

7 - 7 = 0

= 2 x

化简

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

\frac{π}{2} - 4 x + 7 + 2 πn - 7

对同类项分组

= - 4 x + 2 πn + \frac{π}{2} + 7 - 7

7 - 7 = 0

= - 4 x + 2 πn + \frac{π}{2}

2 x = - 4 x + 2 πn + \frac{π}{2}

2 x = - 4 x + 2 πn + \frac{π}{2}

2 x = - 4 x + 2 πn + \frac{π}{2}

将

4 x

para o lado esquerdo

2 x = - 4 x + 2 πn + \frac{π}{2}

两边加上

4 x

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

化简

6 x = 2 πn + \frac{π}{2}

6 x = 2 πn + \frac{π}{2}

两边除以

6

6 x = 2 πn + \frac{π}{2}

两边除以

6

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

化简

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

化简

\frac{6 x}{6} : x

\frac{6 x}{6}

数字相除:

\frac{6}{6} = 1

= x

化简

\frac{2 πn}{6} + \frac{\frac{π}{2}}{6} : \frac{4 πn + π}{12}

\frac{2 πn}{6} + \frac{\frac{π}{2}}{6}

使用法则

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

= \frac{2 πn + \frac{π}{2}}{6}

化简

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

2 πn + \frac{π}{2}

将项转换为分式:

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

数字相乘:

2 \cdot 2 = 4

= \frac{4 πn + π}{2}

= \frac{\frac{4 πn + π}{2}}{6}

使用分式法则:

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

= \frac{4 πn + π}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4 πn + π}{12}

x = \frac{4 πn + π}{12}

x = \frac{4 πn + π}{12}

x = \frac{4 πn + π}{12}

2 x + 7 = π - (\frac{π}{2} - (4 x - 7)) + 2 πn : x = - \frac{π + 4 πn - 28}{4}

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

展开

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

π - (\frac{π}{2} - (4 x - 7)) + 2 πn

- (4 x - 7) : - 4 x + 7

- (4 x - 7)

打开括号

= - (4 x) - (- 7)

使用加减运算法则

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

= - 4 x + 7

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

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

- (\frac{π}{2} - 4 x + 7)

打开括号

= - (\frac{π}{2}) - (- 4 x) - (7)

使用加减运算法则

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

= - \frac{π}{2} + 4 x - 7

= π - \frac{π}{2} + 4 x - 7 + 2 πn

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

将

7

到右边

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

两边减去

7

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

化简

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

化简

2 x + 7 - 7 : 2 x

2 x + 7 - 7

同类项相加:

7 - 7 = 0

= 2 x

化简

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

π - \frac{π}{2} + 4 x - 7 + 2 πn - 7

对同类项分组

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

数字相减:

- 7 - 7 = - 14

= 4 x + 2 πn + π - 14 - \frac{π}{2}

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

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

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

将

4 x

para o lado esquerdo

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

两边减去

4 x

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

化简

- 2 x = 2 πn + π - 14 - \frac{π}{2}

- 2 x = 2 πn + π - 14 - \frac{π}{2}

两边除以

- 2

- 2 x = 2 πn + π - 14 - \frac{π}{2}

两边除以

- 2

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

化简

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

化简

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

\frac{- 2 x}{- 2}

使用分式法则:

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

= \frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 πn}{- 2} + \frac{π}{- 2} - \frac{14}{- 2} - \frac{\frac{π}{2}}{- 2} : - \frac{π + 4 πn - 28}{4}

\frac{2 πn}{- 2} + \frac{π}{- 2} - \frac{14}{- 2} - \frac{\frac{π}{2}}{- 2}

使用法则

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

= \frac{2 πn + π - 14 - \frac{π}{2}}{- 2}

使用分式法则:

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

= - \frac{2 πn + π - 14 - \frac{π}{2}}{2}

化简

2 πn + π - 14 - \frac{π}{2} : \frac{π + 4 πn - 28}{2}

2 πn + π - 14 - \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, π = \frac{π 2}{2}, 14 = \frac{14 \cdot 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{π 2}{2} - \frac{14 \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{2 πn \cdot 2 + π 2 - 14 \cdot 2 - π}{2}

2 πn \cdot 2 + π 2 - 14 \cdot 2 - π = π + 4 πn - 28

2 πn \cdot 2 + π 2 - 14 \cdot 2 - π

对同类项分组

= 2 π - π + 2 \cdot 2 πn - 14 \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn - 14 \cdot 2

数字相乘:

2 \cdot 2 = 4

= π + 4 πn - 14 \cdot 2

数字相乘:

14 \cdot 2 = 28

= π + 4 πn - 28

= \frac{π + 4 πn - 28}{2}

= - \frac{\frac{π + 4 πn - 28}{2}}{2}

化简

\frac{\frac{π + 4 πn - 28}{2}}{2} : \frac{π + 4 πn - 28}{4}

\frac{\frac{π + 4 πn - 28}{2}}{2}

使用分式法则:

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

= \frac{π + 4 πn - 28}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn - 28}{4}

= - \frac{π + 4 πn - 28}{4}

x = - \frac{π + 4 πn - 28}{4}

x = - \frac{π + 4 πn - 28}{4}

x = - \frac{π + 4 πn - 28}{4}

x = \frac{4 πn + π}{12}, x = - \frac{π + 4 πn - 28}{4}

x = \frac{4 πn + π}{12}, x = - \frac{π + 4 πn - 28}{4}

作图

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