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受欢迎的三角函数 >

sin(7x)=cos(2x)

解答

sin (7 x) = cos (2 x)

解答

x = \frac{π + 4 πn}{18}, x = \frac{π + 4 πn}{10}

+1

度数

x = 1 0^{\circ} + 4 0^{\circ} n, x = 1 8^{\circ} + 7 2^{\circ} n

求解步骤

sin (7 x) = cos (2 x)

使用三角恒等式改写

sin (7 x) = cos (2 x)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (7 x) = sin (\frac{π}{2} - 2 x)

sin (7 x) = sin (\frac{π}{2} - 2 x)

使用反三角函数性质

sin (7 x) = sin (\frac{π}{2} - 2 x)

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

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

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

7 x = \frac{π}{2} - 2 x + 2 πn : x = \frac{π + 4 πn}{18}

7 x = \frac{π}{2} - 2 x + 2 πn

将

2 x

para o lado esquerdo

7 x = \frac{π}{2} - 2 x + 2 πn

两边加上

2 x

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

化简

9 x = \frac{π}{2} + 2 πn

9 x = \frac{π}{2} + 2 πn

两边除以

9

9 x = \frac{π}{2} + 2 πn

两边除以

9

\frac{9 x}{9} = \frac{\frac{π}{2}}{9} + \frac{2 πn}{9}

化简

\frac{9 x}{9} = \frac{\frac{π}{2}}{9} + \frac{2 πn}{9}

化简

\frac{9 x}{9} : x

\frac{9 x}{9}

数字相除:

\frac{9}{9} = 1

= x

化简

\frac{\frac{π}{2}}{9} + \frac{2 πn}{9} : \frac{π + 4 πn}{18}

\frac{\frac{π}{2}}{9} + \frac{2 πn}{9}

使用法则

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

= \frac{\frac{π}{2} + 2 πn}{9}

化简

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

\frac{π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

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

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{9}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{π + 4 πn}{18}

x = \frac{π + 4 πn}{18}

x = \frac{π + 4 πn}{18}

x = \frac{π + 4 πn}{18}

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

7 x = π - (\frac{π}{2} - 2 x) + 2 πn

展开

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

π - (\frac{π}{2} - 2 x) + 2 πn

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

- (\frac{π}{2} - 2 x)

打开括号

= - (\frac{π}{2}) - (- 2 x)

使用加减运算法则

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

= - \frac{π}{2} + 2 x

= π - \frac{π}{2} + 2 x + 2 πn

7 x = π - \frac{π}{2} + 2 x + 2 πn

将

2 x

para o lado esquerdo

7 x = π - \frac{π}{2} + 2 x + 2 πn

两边减去

2 x

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

化简

5 x = π - \frac{π}{2} + 2 πn

5 x = π - \frac{π}{2} + 2 πn

两边除以

5

5 x = π - \frac{π}{2} + 2 πn

两边除以

5

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

化简

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

化简

\frac{5 x}{5} : x

\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{π + 4 πn}{10}

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

使用法则

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

= \frac{π - \frac{π}{2} + 2 πn}{5}

化简

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

π - \frac{π}{2} + 2 πn

将项转换为分式:

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

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

π 2 - π + 2 πn \cdot 2 = π + 4 πn

π 2 - π + 2 πn \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= π + 4 πn

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{5}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{18}, x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{18}, x = \frac{π + 4 πn}{10}

作图

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