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受欢迎的三角函数 >

sin(2x)-sin(4x)+sin(6x)=0

解答

sin (2 x) - sin (4 x) + sin (6 x) = 0

解答

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π + 6 πn}{6}, x = \frac{5 π + 6 πn}{6}, x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n, x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

sin (2 x) - sin (4 x) + sin (6 x) = 0

令

: u = 2 x

sin (u) - sin (2 u) + sin (3 u) = 0

使用三角恒等式改写

- sin (2 u) + sin (3 u) + sin (u)

使用倍角公式:

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

= - 2 sin (u) cos (u) + sin (3 u) + sin (u)

sin (3 u) = 3 sin (u) - 4 sin^{3} (u)

sin (3 u)

使用三角恒等式改写

sin (3 u)

改写为

= sin (2 u + u)

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

使用倍角公式:

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

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

化简

cos (2 u) sin (u) + cos (u) \cdot 2 sin (u) cos (u) : sin (u) cos (2 u) + 2 cos^{2} (u) sin (u)

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

cos (u) \cdot 2 sin (u) cos (u) = 2 cos^{2} (u) sin (u)

cos (u) 2 sin (u) cos (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (u) cos (u) = cos^{1 + 1} (u)

= 2 sin (u) cos^{1 + 1} (u)

数字相加:

1 + 1 = 2

= 2 sin (u) cos^{2} (u)

= sin (u) cos (2 u) + 2 cos^{2} (u) sin (u)

= sin (u) cos (2 u) + 2 cos^{2} (u) sin (u)

= sin (u) cos (2 u) + 2 cos^{2} (u) sin (u)

使用倍角公式:

cos (2 u) = 1 - 2 sin^{2} (u)

= (1 - 2 sin^{2} (u)) sin (u) + 2 cos^{2} (u) sin (u)

使用毕达哥拉斯恒等式:

cos^{2} (u) + sin^{2} (u) = 1

cos^{2} (u) = 1 - sin^{2} (u)

= (1 - 2 sin^{2} (u)) sin (u) + 2 (1 - sin^{2} (u)) sin (u)

乘开

(1 - 2 sin^{2} (u)) sin (u) + 2 (1 - sin^{2} (u)) sin (u) : - 4 sin^{3} (u) + 3 sin (u)

(1 - 2 sin^{2} (u)) sin (u) + 2 (1 - sin^{2} (u)) sin (u)

= sin (u) (1 - 2 sin^{2} (u)) + 2 sin (u) (1 - sin^{2} (u))

乘开

sin (u) (1 - 2 sin^{2} (u)) : sin (u) - 2 sin^{3} (u)

sin (u) (1 - 2 sin^{2} (u))

使用分配律:

a (b - c) = ab - a c

a = sin (u), b = 1, c = 2 sin^{2} (u)

= sin (u) 1 - sin (u) 2 sin^{2} (u)

= 1 sin (u) - 2 sin^{2} (u) sin (u)

化简

1 \cdot sin (u) - 2 sin^{2} (u) sin (u) : sin (u) - 2 sin^{3} (u)

1 sin (u) - 2 sin^{2} (u) sin (u)

1 \cdot sin (u) = sin (u)

1 sin (u)

乘以:

1 \cdot sin (u) = sin (u)

= sin (u)

2 sin^{2} (u) sin (u) = 2 sin^{3} (u)

2 sin^{2} (u) sin (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin^{2} (u) sin (u) = sin^{2 + 1} (u)

= 2 sin^{2 + 1} (u)

数字相加:

2 + 1 = 3

= 2 sin^{3} (u)

= sin (u) - 2 sin^{3} (u)

= sin (u) - 2 sin^{3} (u)

= sin (u) - 2 sin^{3} (u) + 2 (1 - sin^{2} (u)) sin (u)

乘开

2 sin (u) (1 - sin^{2} (u)) : 2 sin (u) - 2 sin^{3} (u)

2 sin (u) (1 - sin^{2} (u))

使用分配律:

a (b - c) = ab - a c

a = 2 sin (u), b = 1, c = sin^{2} (u)

= 2 sin (u) 1 - 2 sin (u) sin^{2} (u)

= 2 \cdot 1 sin (u) - 2 sin^{2} (u) sin (u)

化简

2 \cdot 1 \cdot sin (u) - 2 sin^{2} (u) sin (u) : 2 sin (u) - 2 sin^{3} (u)

2 \cdot 1 sin (u) - 2 sin^{2} (u) sin (u)

2 \cdot 1 \cdot sin (u) = 2 sin (u)

2 \cdot 1 sin (u)

数字相乘:

2 \cdot 1 = 2

= 2 sin (u)

2 sin^{2} (u) sin (u) = 2 sin^{3} (u)

2 sin^{2} (u) sin (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin^{2} (u) sin (u) = sin^{2 + 1} (u)

= 2 sin^{2 + 1} (u)

数字相加:

2 + 1 = 3

= 2 sin^{3} (u)

= 2 sin (u) - 2 sin^{3} (u)

= 2 sin (u) - 2 sin^{3} (u)

= sin (u) - 2 sin^{3} (u) + 2 sin (u) - 2 sin^{3} (u)

化简

sin (u) - 2 sin^{3} (u) + 2 sin (u) - 2 sin^{3} (u) : - 4 sin^{3} (u) + 3 sin (u)

sin (u) - 2 sin^{3} (u) + 2 sin (u) - 2 sin^{3} (u)

对同类项分组

= - 2 sin^{3} (u) - 2 sin^{3} (u) + sin (u) + 2 sin (u)

同类项相加:

- 2 sin^{3} (u) - 2 sin^{3} (u) = - 4 sin^{3} (u)

= - 4 sin^{3} (u) + sin (u) + 2 sin (u)

同类项相加:

sin (u) + 2 sin (u) = 3 sin (u)

= - 4 sin^{3} (u) + 3 sin (u)

= - 4 sin^{3} (u) + 3 sin (u)

= - 4 sin^{3} (u) + 3 sin (u)

= 3 sin (u) - 4 sin^{3} (u) + sin (u) - 2 cos (u) sin (u)

化简

= 4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u)

4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u) = 0

分解

4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u) : 2 sin (u) (2 - 2 sin^{2} (u) - cos (u))

4 sin (u) - 4 sin^{3} (u) - 2 cos (u) sin (u)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{3} (u) = sin (u) sin^{2} (u)

= 4 sin (u) - 4 sin (u) sin^{2} (u) - 2 sin (u) cos (u)

将

- 4

改写为

2 \cdot 2

将

4

改写为

2 \cdot 2

= 2 \cdot 2 sin (u) + 2 \cdot 2 sin (u) sin^{2} (u) - 2 sin (u) cos (u)

因式分解出通项

2 sin (u)

= 2 sin (u) (2 - 2 sin^{2} (u) - cos (u))

2 sin (u) (2 - 2 sin^{2} (u) - cos (u)) = 0

分别求解每个部分

sin (u) = 0 or 2 - 2 sin^{2} (u) - cos (u) = 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

2 - 2 sin^{2} (u) - cos (u) = 0 : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

2 - 2 sin^{2} (u) - cos (u) = 0

使用三角恒等式改写

2 - cos (u) - 2 sin^{2} (u)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 2 - cos (u) - 2 (1 - cos^{2} (u))

化简

2 - cos (u) - 2 (1 - cos^{2} (u)) : 2 cos^{2} (u) - cos (u)

2 - cos (u) - 2 (1 - cos^{2} (u))

乘开

- 2 (1 - cos^{2} (u)) : - 2 + 2 cos^{2} (u)

- 2 (1 - cos^{2} (u))

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (u)

= - 2 \cdot 1 - (- 2) cos^{2} (u)

使用加减运算法则

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (u)

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (u)

= 2 - cos (u) - 2 + 2 cos^{2} (u)

化简

2 - cos (u) - 2 + 2 cos^{2} (u) : 2 cos^{2} (u) - cos (u)

2 - cos (u) - 2 + 2 cos^{2} (u)

对同类项分组

= - cos (u) + 2 cos^{2} (u) + 2 - 2

2 - 2 = 0

= 2 cos^{2} (u) - cos (u)

= 2 cos^{2} (u) - cos (u)

= 2 cos^{2} (u) - cos (u)

- cos (u) + 2 cos^{2} (u) = 0

用替代法求解

- cos (u) + 2 cos^{2} (u) = 0

令

: cos (u) = u

- u + 2 u^{2} = 0

- u + 2 u^{2} = 0 : u = \frac{1}{2}, u = 0

- u + 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} - u = 0

使用求根公式求解

2 u^{2} - u = 0

二次方程求根公式:

若

a = 2, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 2 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 1}{2 \cdot 2}

u = \frac{- ( - 1 ) + 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 1 ) + 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 + 1}{2 \cdot 2}

数字相加:

1 + 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- ( - 1 ) - 1}{2 \cdot 2} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 - 1}{2 \cdot 2}

数字相减:

1 - 1 = 0

= \frac{0}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = \frac{1}{2}, u = 0

u = cos (u)

代回

cos (u) = \frac{1}{2}, cos (u) = 0

cos (u) = \frac{1}{2}, cos (u) = 0

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

cos (u) = \frac{1}{2}

cos (u) = \frac{1}{2}

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}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

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

合并所有解

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

合并所有解

u = 2 πn, u = π + 2 πn, u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 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 = \frac{π}{3} + 2 πn : x = \frac{π + 6 πn}{6}

2 x = \frac{π}{3} + 2 πn

两边除以

2

2 x = \frac{π}{3} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{π}{3}}{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{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π + 6 πn}{6}

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

使用法则

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

= \frac{\frac{π}{3} + 2 πn}{2}

化简

\frac{π}{3} + 2 πn : \frac{π + 6 πn}{3}

\frac{π}{3} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 3}{3}

= \frac{π}{3} + \frac{2 πn \cdot 3}{3}

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 3}{3}

数字相乘:

2 \cdot 3 = 6

= \frac{π + 6 πn}{3}

= \frac{\frac{π + 6 πn}{3}}{2}

使用分式法则:

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

= \frac{π + 6 πn}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π + 6 πn}{6}

x = \frac{π + 6 πn}{6}

x = \frac{π + 6 πn}{6}

x = \frac{π + 6 πn}{6}

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π + 6 πn}{6}

2 x = \frac{5 π}{3} + 2 πn

两边除以

2

2 x = \frac{5 π}{3} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{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{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π + 6 πn}{6}

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{\frac{5 π}{3} + 2 πn}{2}

化简

\frac{5 π}{3} + 2 πn : \frac{5 π + 6 πn}{3}

\frac{5 π}{3} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 3}{3}

= \frac{5 π}{3} + \frac{2 πn \cdot 3}{3}

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{5 π + 2 πn \cdot 3}{3}

数字相乘:

2 \cdot 3 = 6

= \frac{5 π + 6 πn}{3}

= \frac{\frac{5 π + 6 πn}{3}}{2}

使用分式法则:

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

= \frac{5 π + 6 πn}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{5 π + 6 πn}{6}

x = \frac{5 π + 6 πn}{6}

x = \frac{5 π + 6 πn}{6}

x = \frac{5 π + 6 πn}{6}

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

2 x = \frac{π}{2} + 2 πn

两边除以

2

2 x = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

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

使用法则

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

= \frac{\frac{π}{2} + 2 πn}{2}

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}}{2}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{4}

x = \frac{π + 4 πn}{4}

x = \frac{π + 4 πn}{4}

x = \frac{π + 4 πn}{4}

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

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

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

使用法则

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

= \frac{\frac{3 π}{2} + 2 πn}{2}

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}}{2}

使用分式法则:

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

= \frac{3 π + 4 πn}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

x = \frac{3 π + 4 πn}{4}

x = πn, x = \frac{π + 2 πn}{2}, x = \frac{π + 6 πn}{6}, x = \frac{5 π + 6 πn}{6}, x = \frac{π + 4 πn}{4}, x = \frac{3 π + 4 πn}{4}

作图

流行的例子

sin(x)sin(2x)=0

3sin(x)=3sin(2x)