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受欢迎的三角函数 >

sin(8x)-2cos(4x)=0

解答

sin (8 x) - 2 cos (4 x) = 0

解答

x = \frac{π + 4 πn}{8}, x = \frac{3 π + 4 πn}{8}

+1

度数

x = 22. 5^{\circ} + 9 0^{\circ} n, x = 67. 5^{\circ} + 9 0^{\circ} n

求解步骤

sin (8 x) - 2 cos (4 x) = 0

令

: u = 4 x

sin (2 u) - 2 cos (u) = 0

使用三角恒等式改写

sin (2 u) - 2 cos (u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (u) cos (u) - 2 cos (u)

- 2 cos (u) + 2 cos (u) sin (u) = 0

分解

- 2 cos (u) + 2 cos (u) sin (u) : 2 cos (u) (sin (u) - 1)

- 2 cos (u) + 2 cos (u) sin (u)

因式分解出通项

2 cos (u)

= 2 cos (u) (- 1 + sin (u))

2 cos (u) (sin (u) - 1) = 0

分别求解每个部分

cos (u) = 0 or sin (u) - 1 = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

sin (u) - 1 = 0 : u = \frac{π}{2} + 2 πn

sin (u) - 1 = 0

将

1

到右边

sin (u) - 1 = 0

两边加上

1

sin (u) - 1 + 1 = 0 + 1

化简

sin (u) = 1

sin (u) = 1

sin (u) = 1

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = \frac{π}{2} + 2 πn

u = \frac{π}{2} + 2 πn

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = 4 x

代回

4 x = \frac{π}{2} + 2 πn : x = \frac{π + 4 πn}{8}

4 x = \frac{π}{2} + 2 πn

两边除以

4

4 x = \frac{π}{2} + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π + 4 πn}{8}

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

使用法则

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

= \frac{\frac{π}{2} + 2 πn}{4}

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}}{4}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π + 4 πn}{8}

x = \frac{π + 4 πn}{8}

x = \frac{π + 4 πn}{8}

x = \frac{π + 4 πn}{8}

4 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π + 4 πn}{8}

4 x = \frac{3 π}{2} + 2 πn

两边除以

4

4 x = \frac{3 π}{2} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π + 4 πn}{8}

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4}

使用法则

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

= \frac{\frac{3 π}{2} + 2 πn}{4}

化简

\frac{3 π}{2} + 2 πn : \frac{3 π + 4 πn}{2}

\frac{3 π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{3 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{3 π + 2 πn \cdot 2}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{2}

= \frac{\frac{3 π + 4 πn}{2}}{4}

使用分式法则:

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

= \frac{3 π + 4 πn}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{3 π + 4 πn}{8}

x = \frac{π + 4 πn}{8}, x = \frac{3 π + 4 πn}{8}

作图

流行的例子

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1.812 = 316 \cdot cos (1.496 \cdot x) + 1.496

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