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受欢迎的三角函数 >

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

解答

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

解答

x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}, x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

+1

度数

x = - 6 0^{\circ} - 48 0^{\circ} n, x = - 30 0^{\circ} - 48 0^{\circ} n, x = 6 0^{\circ} + 48 0^{\circ} n, x = 30 0^{\circ} + 48 0^{\circ} n

求解步骤

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

两边减去

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

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

使用三角恒等式改写

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

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

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

使用法则

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

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

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

分解

- sin^{2} (\frac{- x + π}{2}) + sin^{2} (x) : (sin (x) + sin (\frac{- x + π}{2})) (sin (x) - sin (\frac{- x + π}{2}))

- sin^{2} (\frac{- x + π}{2}) + sin^{2} (x)

使用平方差公式:

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

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

= (sin (x) + sin (\frac{- x + π}{2})) (sin (x) - sin (\frac{- x + π}{2}))

(sin (x) + sin (\frac{- x + π}{2})) (sin (x) - sin (\frac{- x + π}{2})) = 0

分别求解每个部分

sin (x) + sin (\frac{- x + π}{2}) = 0 or sin (x) - sin (\frac{- x + π}{2}) = 0

sin (x) + sin (\frac{- x + π}{2}) = 0 : x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}, x = 8 πn - π, x = 3 π + 8 πn

sin (x) + sin (\frac{- x + π}{2}) = 0

使用三角恒等式改写

sin (\frac{- x + π}{2}) + sin (x)

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{\frac{- x + π}{2} + x}{2}) cos (\frac{\frac{- x + π}{2} - x}{2})

化简

2 sin (\frac{\frac{- x + π}{2} + x}{2}) cos (\frac{\frac{- x + π}{2} - x}{2}) : 2 sin (\frac{x + π}{4}) cos (\frac{- 3 x + π}{4})

2 sin (\frac{\frac{- x + π}{2} + x}{2}) cos (\frac{\frac{- x + π}{2} - x}{2})

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

\frac{\frac{- x + π}{2} + x}{2}

化简

\frac{- x + π}{2} + x : \frac{x + π}{2}

\frac{- x + π}{2} + x

将项转换为分式:

x = \frac{x 2}{2}

= \frac{- x + π}{2} + \frac{x \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{- x + π + x \cdot 2}{2}

- x + π + x \cdot 2 = x + π

- x + π + x \cdot 2

对同类项分组

= - x + 2 x + π

同类项相加:

- x + 2 x = x

= x + π

= \frac{x + π}{2}

= \frac{\frac{x + π}{2}}{2}

使用分式法则:

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

= \frac{x + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{x + π}{4}

= 2 sin (\frac{x + π}{4}) cos (\frac{\frac{- x + π}{2} - x}{2})

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

\frac{\frac{- x + π}{2} - x}{2}

化简

\frac{- x + π}{2} - x : \frac{- 3 x + π}{2}

\frac{- x + π}{2} - x

将项转换为分式:

x = \frac{x 2}{2}

= \frac{- x + π}{2} - \frac{x \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{- x + π - x \cdot 2}{2}

- x + π - x \cdot 2 = - 3 x + π

- x + π - x \cdot 2

对同类项分组

= - x - 2 x + π

同类项相加:

- x - 2 x = - 3 x

= - 3 x + π

= \frac{- 3 x + π}{2}

= \frac{\frac{- 3 x + π}{2}}{2}

使用分式法则:

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

= \frac{- 3 x + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 3 x + π}{4}

= 2 sin (\frac{x + π}{4}) cos (\frac{- 3 x + π}{4})

= 2 sin (\frac{x + π}{4}) cos (\frac{- 3 x + π}{4})

2 cos (\frac{π - 3 x}{4}) sin (\frac{x + π}{4}) = 0

分别求解每个部分

cos (\frac{π - 3 x}{4}) = 0 or sin (\frac{x + π}{4}) = 0

cos (\frac{π - 3 x}{4}) = 0 : x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}

cos (\frac{π - 3 x}{4}) = 0

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

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

解

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

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

在两边乘以

4

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

在两边乘以

4

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

化简

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

化简

\frac{4 ( π - 3 x )}{4} : π - 3 x

\frac{4 ( π - 3 x )}{4}

数字相除:

\frac{4}{4} = 1

= π - 3 x

化简

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

4 \cdot \frac{π}{2} + 4 \cdot 2 πn

4 \cdot \frac{π}{2} = 2 π

4 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 2 π + 8 πn

π - 3 x = 2 π + 8 πn

π - 3 x = 2 π + 8 πn

π - 3 x = 2 π + 8 πn

将

π

到右边

π - 3 x = 2 π + 8 πn

两边减去

π

π - 3 x - π = 2 π + 8 πn - π

化简

- 3 x = π + 8 πn

- 3 x = π + 8 πn

两边除以

- 3

- 3 x = π + 8 πn

两边除以

- 3

\frac{- 3 x}{- 3} = \frac{π}{- 3} + \frac{8 πn}{- 3}

化简

\frac{- 3 x}{- 3} = \frac{π}{- 3} + \frac{8 πn}{- 3}

化简

\frac{- 3 x}{- 3} : x

\frac{- 3 x}{- 3}

使用分式法则:

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

= \frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{π}{- 3} + \frac{8 πn}{- 3} : - \frac{π}{3} - \frac{8 πn}{3}

\frac{π}{- 3} + \frac{8 πn}{- 3}

使用分式法则:

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

= - \frac{π}{3} + \frac{8 πn}{- 3}

使用分式法则:

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

= - \frac{π}{3} - \frac{8 πn}{3}

x = - \frac{π}{3} - \frac{8 πn}{3}

x = - \frac{π}{3} - \frac{8 πn}{3}

x = - \frac{π}{3} - \frac{8 πn}{3}

解

\frac{π - 3 x}{4} = \frac{3 π}{2} + 2 πn : x = - \frac{5 π}{3} - \frac{8 πn}{3}

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

在两边乘以

4

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

在两边乘以

4

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

化简

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

化简

\frac{4 ( π - 3 x )}{4} : π - 3 x

\frac{4 ( π - 3 x )}{4}

数字相除:

\frac{4}{4} = 1

= π - 3 x

化简

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn : 6 π + 8 πn

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

4 \cdot \frac{3 π}{2} = 6 π

4 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 4}{2}

数字相乘:

3 \cdot 4 = 12

= \frac{12 π}{2}

数字相除:

\frac{12}{2} = 6

= 6 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 6 π + 8 πn

π - 3 x = 6 π + 8 πn

π - 3 x = 6 π + 8 πn

π - 3 x = 6 π + 8 πn

将

π

到右边

π - 3 x = 6 π + 8 πn

两边减去

π

π - 3 x - π = 6 π + 8 πn - π

化简

- 3 x = 5 π + 8 πn

- 3 x = 5 π + 8 πn

两边除以

- 3

- 3 x = 5 π + 8 πn

两边除以

- 3

\frac{- 3 x}{- 3} = \frac{5 π}{- 3} + \frac{8 πn}{- 3}

化简

\frac{- 3 x}{- 3} = \frac{5 π}{- 3} + \frac{8 πn}{- 3}

化简

\frac{- 3 x}{- 3} : x

\frac{- 3 x}{- 3}

使用分式法则:

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

= \frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{5 π}{- 3} + \frac{8 πn}{- 3} : - \frac{5 π}{3} - \frac{8 πn}{3}

\frac{5 π}{- 3} + \frac{8 πn}{- 3}

使用分式法则:

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

= - \frac{5 π}{3} + \frac{8 πn}{- 3}

使用分式法则:

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

= - \frac{5 π}{3} - \frac{8 πn}{3}

x = - \frac{5 π}{3} - \frac{8 πn}{3}

x = - \frac{5 π}{3} - \frac{8 πn}{3}

x = - \frac{5 π}{3} - \frac{8 πn}{3}

x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}

sin (\frac{x + π}{4}) = 0 : x = 8 πn - π, x = 3 π + 8 πn

sin (\frac{x + π}{4}) = 0

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

\frac{x + π}{4} = 0 + 2 πn, \frac{x + π}{4} = π + 2 πn

解

\frac{x + π}{4} = 0 + 2 πn : x = 8 πn - π

\frac{x + π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{x + π}{4} = 2 πn

在两边乘以

4

\frac{x + π}{4} = 2 πn

在两边乘以

4

\frac{4 ( x + π )}{4} = 4 \cdot 2 πn

化简

x + π = 8 πn

x + π = 8 πn

将

π

到右边

x + π = 8 πn

两边减去

π

x + π - π = 8 πn - π

化简

x = 8 πn - π

x = 8 πn - π

解

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

\frac{x + π}{4} = π + 2 πn

在两边乘以

4

\frac{x + π}{4} = π + 2 πn

在两边乘以

4

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

化简

x + π = 4 π + 8 πn

x + π = 4 π + 8 πn

将

π

到右边

x + π = 4 π + 8 πn

两边减去

π

x + π - π = 4 π + 8 πn - π

化简

x = 3 π + 8 πn

x = 3 π + 8 πn

x = 8 πn - π, x = 3 π + 8 πn

合并所有解

x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}, x = 8 πn - π, x = 3 π + 8 πn

sin (x) - sin (\frac{- x + π}{2}) = 0 : x = π + 8 πn, x = 5 π + 8 πn, x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

sin (x) - sin (\frac{- x + π}{2}) = 0

使用三角恒等式改写

- sin (\frac{- x + π}{2}) + sin (x)

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{x - \frac{- x + π}{2}}{2}) cos (\frac{x + \frac{- x + π}{2}}{2})

化简

2 sin (\frac{x - \frac{- x + π}{2}}{2}) cos (\frac{x + \frac{- x + π}{2}}{2}) : 2 sin (\frac{3 x - π}{4}) cos (\frac{x + π}{4})

2 sin (\frac{x - \frac{- x + π}{2}}{2}) cos (\frac{x + \frac{- x + π}{2}}{2})

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

\frac{x - \frac{- x + π}{2}}{2}

化简

x - \frac{- x + π}{2} : \frac{3 x - π}{2}

x - \frac{- x + π}{2}

将项转换为分式:

x = \frac{x 2}{2}

= \frac{x \cdot 2}{2} - \frac{- x + π}{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{x \cdot 2 - ( - x + π )}{2}

乘开

x \cdot 2 - (- x + π) : 3 x - π

x \cdot 2 - (- x + π)

= 2 x - (- x + π)

- (- x + π) : x - π

- (- x + π)

打开括号

= - (- x) - (π)

使用加减运算法则

- (- a) = a, - (a) = - a

= x - π

= x \cdot 2 + x - π

同类项相加:

2 x + x = 3 x

= 3 x - π

= \frac{3 x - π}{2}

= \frac{\frac{3 x - π}{2}}{2}

使用分式法则:

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

= \frac{3 x - π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 x - π}{4}

= 2 sin (\frac{3 x - π}{4}) cos (\frac{x + \frac{- x + π}{2}}{2})

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

\frac{x + \frac{- x + π}{2}}{2}

化简

x + \frac{- x + π}{2} : \frac{x + π}{2}

x + \frac{- x + π}{2}

将项转换为分式:

x = \frac{x 2}{2}

= \frac{x \cdot 2}{2} + \frac{- x + π}{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{x \cdot 2 - x + π}{2}

同类项相加:

2 x - x = x

= \frac{x + π}{2}

= \frac{\frac{x + π}{2}}{2}

使用分式法则:

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

= \frac{x + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{x + π}{4}

= 2 sin (\frac{3 x - π}{4}) cos (\frac{x + π}{4})

= 2 sin (\frac{3 x - π}{4}) cos (\frac{x + π}{4})

2 cos (\frac{x + π}{4}) sin (\frac{- π + 3 x}{4}) = 0

分别求解每个部分

cos (\frac{x + π}{4}) = 0 or sin (\frac{- π + 3 x}{4}) = 0

cos (\frac{x + π}{4}) = 0 : x = π + 8 πn, x = 5 π + 8 πn

cos (\frac{x + π}{4}) = 0

cos (\frac{x + π}{4}) = 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{x + π}{4} = \frac{π}{2} + 2 πn, \frac{x + π}{4} = \frac{3 π}{2} + 2 πn

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

解

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

\frac{x + π}{4} = \frac{π}{2} + 2 πn

在两边乘以

4

\frac{x + π}{4} = \frac{π}{2} + 2 πn

在两边乘以

4

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

化简

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

化简

\frac{4 ( x + π )}{4} : x + π

\frac{4 ( x + π )}{4}

数字相除:

\frac{4}{4} = 1

= x + π

化简

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

4 \cdot \frac{π}{2} + 4 \cdot 2 πn

4 \cdot \frac{π}{2} = 2 π

4 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 2 π + 8 πn

x + π = 2 π + 8 πn

x + π = 2 π + 8 πn

x + π = 2 π + 8 πn

将

π

到右边

x + π = 2 π + 8 πn

两边减去

π

x + π - π = 2 π + 8 πn - π

化简

x = π + 8 πn

x = π + 8 πn

解

\frac{x + π}{4} = \frac{3 π}{2} + 2 πn : x = 5 π + 8 πn

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

在两边乘以

4

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

在两边乘以

4

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

化简

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

化简

\frac{4 ( x + π )}{4} : x + π

\frac{4 ( x + π )}{4}

数字相除:

\frac{4}{4} = 1

= x + π

化简

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn : 6 π + 8 πn

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

4 \cdot \frac{3 π}{2} = 6 π

4 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 4}{2}

数字相乘:

3 \cdot 4 = 12

= \frac{12 π}{2}

数字相除:

\frac{12}{2} = 6

= 6 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 6 π + 8 πn

x + π = 6 π + 8 πn

x + π = 6 π + 8 πn

x + π = 6 π + 8 πn

将

π

到右边

x + π = 6 π + 8 πn

两边减去

π

x + π - π = 6 π + 8 πn - π

化简

x = 5 π + 8 πn

x = 5 π + 8 πn

x = π + 8 πn, x = 5 π + 8 πn

sin (\frac{- π + 3 x}{4}) = 0 : x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

sin (\frac{- π + 3 x}{4}) = 0

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

\frac{- π + 3 x}{4} = 0 + 2 πn, \frac{- π + 3 x}{4} = π + 2 πn

解

\frac{- π + 3 x}{4} = 0 + 2 πn : x = \frac{8 πn}{3} + \frac{π}{3}

\frac{- π + 3 x}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{- π + 3 x}{4} = 2 πn

在两边乘以

4

\frac{- π + 3 x}{4} = 2 πn

在两边乘以

4

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

化简

- π + 3 x = 8 πn

- π + 3 x = 8 πn

将

π

到右边

- π + 3 x = 8 πn

两边加上

π

- π + 3 x + π = 8 πn + π

化简

3 x = 8 πn + π

3 x = 8 πn + π

两边除以

3

3 x = 8 πn + π

两边除以

3

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

化简

x = \frac{8 πn}{3} + \frac{π}{3}

x = \frac{8 πn}{3} + \frac{π}{3}

解

\frac{- π + 3 x}{4} = π + 2 πn : x = \frac{5 π}{3} + \frac{8 πn}{3}

\frac{- π + 3 x}{4} = π + 2 πn

在两边乘以

4

\frac{- π + 3 x}{4} = π + 2 πn

在两边乘以

4

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

化简

- π + 3 x = 4 π + 8 πn

- π + 3 x = 4 π + 8 πn

将

π

到右边

- π + 3 x = 4 π + 8 πn

两边加上

π

- π + 3 x + π = 4 π + 8 πn + π

化简

3 x = 5 π + 8 πn

3 x = 5 π + 8 πn

两边除以

3

3 x = 5 π + 8 πn

两边除以

3

\frac{3 x}{3} = \frac{5 π}{3} + \frac{8 πn}{3}

化简

x = \frac{5 π}{3} + \frac{8 πn}{3}

x = \frac{5 π}{3} + \frac{8 πn}{3}

x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

合并所有解

x = π + 8 πn, x = 5 π + 8 πn, x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

合并所有解

x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}, x = 8 πn - π, x = 3 π + 8 πn, x = π + 8 πn, x = 5 π + 8 πn, x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

合并重叠的区间

x = - \frac{π}{3} - \frac{8 πn}{3}, x = - \frac{5 π}{3} - \frac{8 πn}{3}, x = \frac{8 πn}{3} + \frac{π}{3}, x = \frac{5 π}{3} + \frac{8 πn}{3}

作图

流行的例子

solvefor x,sin(x)=y

cos(2x)=1-2sin(x)

sin(θ-pi/3)-1=0