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受欢迎的三角函数 >

2sin(4x)=sin(2x)

解答

2 sin (4 x) = sin (2 x)

解答

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{1.31811 \dots + 2 πn}{2}, x = \frac{2 π - 1.31811 \dots + 2 πn}{2}

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n, x = 37.76124 \dots^{\circ} + 18 0^{\circ} n, x = 142.23875 \dots^{\circ} + 18 0^{\circ} n

求解步骤

2 sin (4 x) = sin (2 x)

两边减去

sin (2 x)

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

令

: u = 2 x

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

使用三角恒等式改写

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

使用倍角公式:

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

= - sin (u) + 2 \cdot 2 sin (u) cos (u)

化简

= - sin (u) + 4 sin (u) cos (u)

- sin (u) + 4 cos (u) sin (u) = 0

分解

- sin (u) + 4 cos (u) sin (u) : sin (u) (4 cos (u) - 1)

- sin (u) + 4 cos (u) sin (u)

因式分解出通项

sin (u)

= sin (u) (- 1 + 4 cos (u))

sin (u) (4 cos (u) - 1) = 0

分别求解每个部分

sin (u) = 0 or 4 cos (u) - 1 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 0

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 = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

4 cos (u) - 1 = 0 : u = arccos (\frac{1}{4}) + 2 πn, u = 2 π - arccos (\frac{1}{4}) + 2 πn

4 cos (u) - 1 = 0

将

1

到右边

4 cos (u) - 1 = 0

两边加上

1

4 cos (u) - 1 + 1 = 0 + 1

化简

4 cos (u) = 1

4 cos (u) = 1

两边除以

4

4 cos (u) = 1

两边除以

4

\frac{4 cos ( u )}{4} = \frac{1}{4}

化简

cos (u) = \frac{1}{4}

cos (u) = \frac{1}{4}

使用反三角函数性质

cos (u) = \frac{1}{4}

cos (u) = \frac{1}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

u = arccos (\frac{1}{4}) + 2 πn, u = 2 π - arccos (\frac{1}{4}) + 2 πn

u = arccos (\frac{1}{4}) + 2 πn, u = 2 π - arccos (\frac{1}{4}) + 2 πn

合并所有解

u = 2 πn, u = π + 2 πn, u = arccos (\frac{1}{4}) + 2 πn, u = 2 π - arccos (\frac{1}{4}) + 2 πn

u = 2 x

代回

2 x = 2 πn : x = πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

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

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{π}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

2 x = arccos (\frac{1}{4}) + 2 πn : x = \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

2 x = arccos (\frac{1}{4}) + 2 πn

两边除以

2

2 x = arccos (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2} : \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

\frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}

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

2 x = 2 π - arccos (\frac{1}{4}) + 2 πn

两边除以

2

2 x = 2 π - arccos (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{2 π - arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{2 π - arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{2 π - arccos ( \frac{1}{4} ) + 2 πn}{2}

x = \frac{2 π - arccos ( \frac{1}{4} ) + 2 πn}{2}

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{arccos ( \frac{1}{4} ) + 2 πn}{2}, x = \frac{2 π - arccos ( \frac{1}{4} ) + 2 πn}{2}

以小数形式表示解

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{1.31811 \dots + 2 πn}{2}, x = \frac{2 π - 1.31811 \dots + 2 πn}{2}

作图

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