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受欢迎的三角函数 >

2-2sin^2(x)=2cos^2(x/2)

解答

2 - 2 sin^{2} (x) = 2 cos^{2} (\frac{x}{2})

解答

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

+1

度数

x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n, x = 24 0^{\circ} + 72 0^{\circ} n, x = 48 0^{\circ} + 72 0^{\circ} n, x = 12 0^{\circ} + 48 0^{\circ} n, x = 36 0^{\circ} + 48 0^{\circ} n, x = 0^{\circ} + 144 0^{\circ} n, x = 72 0^{\circ} + 144 0^{\circ} n

求解步骤

2 - 2 sin^{2} (x) = 2 cos^{2} (\frac{x}{2})

两边减去

2 cos^{2} (\frac{x}{2})

2 - 2 sin^{2} (x) - 2 cos^{2} (\frac{x}{2}) = 0

令

: u = \frac{x}{2}

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

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

2 - 2 = 0

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

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

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

分解

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

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

因式分解出通项

2

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

分解

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

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

使用平方差公式:

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

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

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

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

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

分别求解每个部分

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

sin (u) + sin (2 u) = 0 : u = 2 πn, u = π + 2 πn, u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

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

使用三角恒等式改写

sin (2 u) + sin (u)

使用倍角公式:

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

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

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

分解

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

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

因式分解出通项

sin (u)

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

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

分别求解每个部分

sin (u) = 0 or 2 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

2 cos (u) + 1 = 0 : u = \frac{2 π}{3} + 2 πn, u = \frac{4 π}{3} + 2 πn

2 cos (u) + 1 = 0

将

1

到右边

2 cos (u) + 1 = 0

两边减去

1

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

化简

2 cos (u) = - 1

2 cos (u) = - 1

两边除以

2

2 cos (u) = - 1

两边除以

2

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

化简

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

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

合并所有解

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

sin (u) - sin (2 u) = 0 : u = \frac{π}{3} + \frac{4 πn}{3}, u = π + \frac{4 πn}{3}, u = 4 πn, u = 2 π + 4 πn

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

使用三角恒等式改写

- sin (2 u) + sin (u)

使用和差化积恒等式:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

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

化简

2 sin (\frac{u - 2 u}{2}) cos (\frac{u + 2 u}{2}) : - 2 cos (\frac{3 u}{2}) sin (\frac{u}{2})

2 sin (\frac{u - 2 u}{2}) cos (\frac{u + 2 u}{2})

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

\frac{u - 2 u}{2}

同类项相加:

u - 2 u = - u

= \frac{- u}{2}

使用分式法则:

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

= - \frac{u}{2}

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

使用负角恒等式:

sin (- x) = - sin (x)

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

去除括号:

(- a) = - a

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

同类项相加:

u + 2 u = 3 u

= - 2 cos (\frac{3 u}{2}) sin (\frac{u}{2})

= - 2 cos (\frac{3 u}{2}) sin (\frac{u}{2})

- 2 cos (\frac{3 u}{2}) sin (\frac{u}{2}) = 0

分别求解每个部分

cos (\frac{3 u}{2}) = 0 or sin (\frac{u}{2}) = 0

cos (\frac{3 u}{2}) = 0 : u = \frac{π}{3} + \frac{4 πn}{3}, u = π + \frac{4 πn}{3}

cos (\frac{3 u}{2}) = 0

cos (\frac{3 u}{2}) = 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}

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

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

解

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

\frac{3 u}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{3 u}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 \cdot 3 u}{2} : 3 u

\frac{2 \cdot 3 u}{2}

数字相乘:

2 \cdot 3 = 6

= \frac{6 u}{2}

数字相除:

\frac{6}{2} = 3

= 3 u

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

3 u = π + 4 πn

3 u = π + 4 πn

3 u = π + 4 πn

两边除以

3

3 u = π + 4 πn

两边除以

3

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

化简

u = \frac{π}{3} + \frac{4 πn}{3}

u = \frac{π}{3} + \frac{4 πn}{3}

解

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

\frac{3 u}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{3 u}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 \cdot 3 u}{2} : 3 u

\frac{2 \cdot 3 u}{2}

数字相乘:

2 \cdot 3 = 6

= \frac{6 u}{2}

数字相除:

\frac{6}{2} = 3

= 3 u

化简

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

3 u = 3 π + 4 πn

3 u = 3 π + 4 πn

3 u = 3 π + 4 πn

两边除以

3

3 u = 3 π + 4 πn

两边除以

3

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

化简

u = π + \frac{4 πn}{3}

u = π + \frac{4 πn}{3}

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

sin (\frac{u}{2}) = 0 : u = 4 πn, u = 2 π + 4 πn

sin (\frac{u}{2}) = 0

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

\frac{u}{2} = 0 + 2 πn, \frac{u}{2} = π + 2 πn

\frac{u}{2} = 0 + 2 πn, \frac{u}{2} = π + 2 πn

解

\frac{u}{2} = 0 + 2 πn : u = 4 πn

\frac{u}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{u}{2} = 2 πn

在两边乘以

2

\frac{u}{2} = 2 πn

在两边乘以

2

\frac{2 u}{2} = 2 \cdot 2 πn

化简

u = 4 πn

u = 4 πn

解

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

\frac{u}{2} = π + 2 πn

在两边乘以

2

\frac{u}{2} = π + 2 πn

在两边乘以

2

\frac{2 u}{2} = 2 π + 2 \cdot 2 πn

化简

u = 2 π + 4 πn

u = 2 π + 4 πn

u = 4 πn, u = 2 π + 4 πn

合并所有解

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

合并所有解

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

u = \frac{x}{2}

代回

\frac{x}{2} = 2 πn : x = 4 πn

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 πn

化简

x = 4 πn

x = 4 πn

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

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

化简

x = 2 π + 4 πn

x = 2 π + 4 πn

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

\frac{x}{2} = \frac{2 π}{3} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{2 π}{3} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{2 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{2 π}{3} = \frac{4 π}{3}

2 \cdot \frac{2 π}{3}

分式相乘:

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

= \frac{2 π 2}{3}

数字相乘:

2 \cdot 2 = 4

= \frac{4 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{4 π}{3} + 4 πn

x = \frac{4 π}{3} + 4 πn

x = \frac{4 π}{3} + 4 πn

x = \frac{4 π}{3} + 4 πn

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

\frac{x}{2} = \frac{4 π}{3} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{4 π}{3} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{4 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{4 π}{3} = \frac{8 π}{3}

2 \cdot \frac{4 π}{3}

分式相乘:

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

= \frac{4 π 2}{3}

数字相乘:

4 \cdot 2 = 8

= \frac{8 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{8 π}{3} + 4 πn

x = \frac{8 π}{3} + 4 πn

x = \frac{8 π}{3} + 4 πn

x = \frac{8 π}{3} + 4 πn

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

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

在两边乘以

2

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

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

2 \cdot \frac{π}{3} = \frac{2 π}{3}

2 \cdot \frac{π}{3}

分式相乘:

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

= \frac{π 2}{3}

2 \cdot \frac{4 πn}{3} = \frac{8 πn}{3}

2 \cdot \frac{4 πn}{3}

分式相乘:

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

= \frac{4 πn \cdot 2}{3}

数字相乘:

4 \cdot 2 = 8

= \frac{8 πn}{3}

= \frac{2 π}{3} + \frac{8 πn}{3}

使用法则

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

= \frac{2 π + 8 πn}{3}

x = \frac{2 π + 8 πn}{3}

x = \frac{2 π + 8 πn}{3}

x = \frac{2 π + 8 πn}{3}

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

\frac{x}{2} = π + \frac{4 πn}{3}

在两边乘以

2

\frac{x}{2} = π + \frac{4 πn}{3}

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 π + 2 \cdot \frac{4 πn}{3}

2 \cdot \frac{4 πn}{3} = \frac{8 πn}{3}

2 \cdot \frac{4 πn}{3}

分式相乘:

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

= \frac{4 πn \cdot 2}{3}

数字相乘:

4 \cdot 2 = 8

= \frac{8 πn}{3}

= 2 π + \frac{8 πn}{3}

x = 2 π + \frac{8 πn}{3}

x = 2 π + \frac{8 πn}{3}

x = 2 π + \frac{8 πn}{3}

\frac{x}{2} = 4 πn : x = 8 πn

\frac{x}{2} = 4 πn

在两边乘以

2

\frac{x}{2} = 4 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 4 πn

化简

x = 8 πn

x = 8 πn

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

\frac{x}{2} = 2 π + 4 πn

在两边乘以

2

\frac{x}{2} = 2 π + 4 πn

在两边乘以

2

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

化简

x = 4 π + 8 πn

x = 4 π + 8 πn

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

合并重叠的区间

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

作图

流行的例子

cos(4x)sin(x)=0

cos (4 x) sin (x) = 0

sqrt(3)cot(pi/2 x)=-3

3 cot (\frac{π}{2} x) = - 3

cos(θ)= 3/5 ,(3pi)/2 <θ<2pi

cos (θ) = \frac{3}{5}, \frac{3 π}{2} < θ < 2 π

cos(2x)=sqrt(3/2)

cos (2 x) = \frac{3}{2}

-3cos(2θ)+2sin(θ)+5=9sin(θ)

- 3 cos (2 θ) + 2 sin (θ) + 5 = 9 sin (θ)