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受欢迎的三角函数 >

cos(2x)+cos(4x)+cos(6x)=0

解答

cos (2 x) + cos (4 x) + cos (6 x) = 0

解答

x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}, x = \frac{3 π + 8 πn}{8}, x = \frac{5 π + 8 πn}{8}, x = \frac{π + 8 πn}{8}, x = \frac{7 π + 8 πn}{8}

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 67. 5^{\circ} + 18 0^{\circ} n, x = 112. 5^{\circ} + 18 0^{\circ} n, x = 22. 5^{\circ} + 18 0^{\circ} n, x = 157. 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 x) + cos (4 x) + cos (6 x) = 0

令

: u = 2 x

cos (u) + cos (2 u) + cos (3 u) = 0

使用三角恒等式改写

cos (2 u) + cos (3 u) + cos (u)

使用倍角公式:

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

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

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

cos (3 u)

使用三角恒等式改写

cos (3 u)

改写为

= cos (2 u + u)

使用角和恒等式:

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

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

使用倍角公式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

乘开

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

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

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

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (u)

1 \cdot cos (u) = cos (u)

1 cos (u)

乘以:

1 \cdot cos (u) = cos (u)

= cos (u)

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

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

= 2 cos^{3} (u) - cos (u) - 2 (1 - cos^{2} (u)) cos (u)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

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

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

2 \cdot 1 cos (u)

数字相乘:

2 \cdot 1 = 2

= 2 cos (u)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (u)

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

同类项相加:

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

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

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

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

= - 1 + 4 cos^{3} (u) - 3 cos (u) + cos (u) + 2 cos^{2} (u)

化简

= - 1 + 4 cos^{3} (u) - 2 cos (u) + 2 cos^{2} (u)

- 1 - 2 cos (u) + 2 cos^{2} (u) + 4 cos^{3} (u) = 0

用替代法求解

- 1 - 2 cos (u) + 2 cos^{2} (u) + 4 cos^{3} (u) = 0

令

: cos (u) = u

- 1 - 2 u + 2 u^{2} + 4 u^{3} = 0

- 1 - 2 u + 2 u^{2} + 4 u^{3} = 0 : u = - \frac{1}{2}, u = - \frac{2}{2}, u = \frac{2}{2}

- 1 - 2 u + 2 u^{2} + 4 u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 u^{3} + 2 u^{2} - 2 u - 1 = 0

因式分解

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

4 u^{3} + 2 u^{2} - 2 u - 1

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

从

- 2 u - 1

分解出因式

- 1

:

- (2 u + 1)

- 2 u - 1

因式分解出通项

- 1

= - (2 u + 1)

从

4 u^{3} + 2 u^{2}

分解出因式

2 u^{2}

:

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

4 u^{3} + 2 u^{2}

使用指数法则:

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

u^{3} = u u^{2}

= 4 u u^{2} + 2 u^{2}

将

4

改写为

2 \cdot 2

= 2 \cdot 2 u u^{2} + 2 u^{2}

因式分解出通项

2 u^{2}

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

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

因式分解出通项

2 u + 1

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

分解

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

2 u^{2} - 1

将

2 u^{2} - 1

改写为

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

2 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

(2 u + 1) (2 u + 1) (2 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

2 u + 1 = 0 or 2 u + 1 = 0 or 2 u - 1 = 0

解

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

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

u = - \frac{1}{2}

u = - \frac{1}{2}

解

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

2 u + 1 = 0

将

1

到右边

2 u + 1 = 0

两边减去

1

2 u + 1 - 1 = 0 - 1

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

解

2 u - 1 = 0 : u = \frac{2}{2}

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

解为

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

u = cos (u)

代回

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

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

cos (u) = - \frac{1}{2} : u = \frac{2 π}{3} + 2 πn, u = \frac{4 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{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

cos (u) = - \frac{2}{2} : u = \frac{3 π}{4} + 2 πn, u = \frac{5 π}{4} + 2 πn

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

cos (u) = - \frac{2}{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{3 π}{4} + 2 πn, u = \frac{5 π}{4} + 2 πn

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

cos (u) = \frac{2}{2} : u = \frac{π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

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

cos (u) = \frac{2}{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}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

u = \frac{π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

合并所有解

u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn, u = \frac{3 π}{4} + 2 πn, u = \frac{5 π}{4} + 2 πn, u = \frac{π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

u = 2 x

代回

2 x = \frac{2 π}{3} + 2 πn : x = \frac{π ( 3 n + 1 )}{3}

2 x = \frac{2 π}{3} + 2 πn

两边除以

2

2 x = \frac{2 π}{3} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{2 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{2 π}{3}}{2} + \frac{2 πn}{2} : \frac{π ( 3 n + 1 )}{3}

\frac{\frac{2 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{2 π}{3} + 2 πn}{2}

化简

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

\frac{2 π}{3} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 3}{3}

= \frac{2 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{2 π + 2 πn \cdot 3}{3}

数字相乘:

2 \cdot 3 = 6

= \frac{2 π + 6 πn}{3}

= \frac{\frac{2 π + 6 πn}{3}}{2}

使用分式法则:

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

= \frac{2 π + 6 πn}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π + 6 πn}{6}

分解

2 π + 6 πn : 2 π (1 + 3 n)

2 π + 6 πn

改写为

= 1 \cdot 2 π + 3 \cdot 2 πn

因式分解出通项

2 π

= 2 π (1 + 3 n)

= \frac{2 π ( 1 + 3 n )}{6}

约分:

2

= \frac{π ( 3 n + 1 )}{3}

x = \frac{π ( 3 n + 1 )}{3}

x = \frac{π ( 3 n + 1 )}{3}

x = \frac{π ( 3 n + 1 )}{3}

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

2 x = \frac{4 π}{3} + 2 πn

两边除以

2

2 x = \frac{4 π}{3} + 2 πn

两边除以

2

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

化简

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

\frac{\frac{4 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{4 π}{3} + 2 πn}{2}

化简

\frac{4 π}{3} + 2 πn : \frac{4 π + 6 πn}{3}

\frac{4 π}{3} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 3}{3}

= \frac{4 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{4 π + 2 πn \cdot 3}{3}

数字相乘:

2 \cdot 3 = 6

= \frac{4 π + 6 πn}{3}

= \frac{\frac{4 π + 6 πn}{3}}{2}

使用分式法则:

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

= \frac{4 π + 6 πn}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{4 π + 6 πn}{6}

分解

4 π + 6 πn : 2 π (2 + 3 n)

4 π + 6 πn

改写为

= 2 \cdot 2 π + 3 \cdot 2 πn

因式分解出通项

2 π

= 2 π (2 + 3 n)

= \frac{2 π ( 2 + 3 n )}{6}

约分:

2

= \frac{π ( 3 n + 2 )}{3}

x = \frac{π ( 3 n + 2 )}{3}

x = \frac{π ( 3 n + 2 )}{3}

x = \frac{π ( 3 n + 2 )}{3}

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

2 x = \frac{3 π}{4} + 2 πn

两边除以

2

2 x = \frac{3 π}{4} + 2 πn

两边除以

2

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

化简

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

\frac{\frac{3 π}{4}}{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}{4} + 2 πn}{2}

化简

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

\frac{3 π}{4} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 4}{4}

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

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 4}{4}

数字相乘:

2 \cdot 4 = 8

= \frac{3 π + 8 πn}{4}

= \frac{\frac{3 π + 8 πn}{4}}{2}

使用分式法则:

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

= \frac{3 π + 8 πn}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{3 π + 8 πn}{8}

x = \frac{3 π + 8 πn}{8}

x = \frac{3 π + 8 πn}{8}

x = \frac{3 π + 8 πn}{8}

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π + 8 πn}{8}

2 x = \frac{5 π}{4} + 2 πn

两边除以

2

2 x = \frac{5 π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{5 π}{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{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π + 8 πn}{8}

\frac{\frac{5 π}{4}}{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}{4} + 2 πn}{2}

化简

\frac{5 π}{4} + 2 πn : \frac{5 π + 8 πn}{4}

\frac{5 π}{4} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 4}{4}

= \frac{5 π}{4} + \frac{2 πn \cdot 4}{4}

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 4}{4}

数字相乘:

2 \cdot 4 = 8

= \frac{5 π + 8 πn}{4}

= \frac{\frac{5 π + 8 πn}{4}}{2}

使用分式法则:

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

= \frac{5 π + 8 πn}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{5 π + 8 πn}{8}

x = \frac{5 π + 8 πn}{8}

x = \frac{5 π + 8 πn}{8}

x = \frac{5 π + 8 πn}{8}

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

2 x = \frac{π}{4} + 2 πn

两边除以

2

2 x = \frac{π}{4} + 2 πn

两边除以

2

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

化简

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

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

使用法则

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

= \frac{\frac{π}{4} + 2 πn}{2}

化简

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

\frac{π}{4} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 4}{4}

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

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 4}{4}

数字相乘:

2 \cdot 4 = 8

= \frac{π + 8 πn}{4}

= \frac{\frac{π + 8 πn}{4}}{2}

使用分式法则:

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

= \frac{π + 8 πn}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π + 8 πn}{8}

x = \frac{π + 8 πn}{8}

x = \frac{π + 8 πn}{8}

x = \frac{π + 8 πn}{8}

2 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π + 8 πn}{8}

2 x = \frac{7 π}{4} + 2 πn

两边除以

2

2 x = \frac{7 π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{7 π}{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{\frac{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π + 8 πn}{8}

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{\frac{7 π}{4} + 2 πn}{2}

化简

\frac{7 π}{4} + 2 πn : \frac{7 π + 8 πn}{4}

\frac{7 π}{4} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 4}{4}

= \frac{7 π}{4} + \frac{2 πn \cdot 4}{4}

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{7 π + 2 πn \cdot 4}{4}

数字相乘:

2 \cdot 4 = 8

= \frac{7 π + 8 πn}{4}

= \frac{\frac{7 π + 8 πn}{4}}{2}

使用分式法则:

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

= \frac{7 π + 8 πn}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{7 π + 8 πn}{8}

x = \frac{7 π + 8 πn}{8}

x = \frac{7 π + 8 πn}{8}

x = \frac{7 π + 8 πn}{8}

x = \frac{π ( 3 n + 1 )}{3}, x = \frac{π ( 3 n + 2 )}{3}, x = \frac{3 π + 8 πn}{8}, x = \frac{5 π + 8 πn}{8}, x = \frac{π + 8 πn}{8}, x = \frac{7 π + 8 πn}{8}

作图

流行的例子

sqrt(2)cos(x)sin(x)-cos(x)=0

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

6sin(x)=6sin(2x)

6 sin (x) = 6 sin (2 x)

sin(5x)+sin(x)=sqrt(3)cos(2x)

sin (5 x) + sin (x) = 3 cos (2 x)

3 sin (2 x) - 1 = 0

-sec(x)=csc(3.45)

- sec (x) = csc (3.45)