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受欢迎的三角函数 >

sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi

解答

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

解答

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

+1

度数

x = 18 0^{\circ}, x = 24 0^{\circ}, x = 36 0^{\circ}, x = 0^{\circ}, x = 6 0^{\circ}

求解步骤

sin (2 x - 5 \cdot \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}

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}

2 x - 5 \cdot \frac{π}{6} = \frac{7 π}{6} + 2 πn, 2 x - 5 \cdot \frac{π}{6} = \frac{11 π}{6} + 2 πn

2 x - 5 \cdot \frac{π}{6} = \frac{7 π}{6} + 2 πn, 2 x - 5 \cdot \frac{π}{6} = \frac{11 π}{6} + 2 πn

解

2 x - 5 \frac{π}{6} = \frac{7 π}{6} + 2 πn : x = πn + π

2 x - 5 \cdot \frac{π}{6} = \frac{7 π}{6} + 2 πn

乘

5 \cdot \frac{π}{6} : \frac{5 π}{6}

5 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 5}{6}

2 x - \frac{5 π}{6} = \frac{7 π}{6} + 2 πn

将

\frac{5 π}{6}

到右边

2 x - \frac{5 π}{6} = \frac{7 π}{6} + 2 πn

两边加上

\frac{5 π}{6}

2 x - \frac{5 π}{6} + \frac{5 π}{6} = \frac{7 π}{6} + 2 πn + \frac{5 π}{6}

化简

2 x - \frac{5 π}{6} + \frac{5 π}{6} = \frac{7 π}{6} + 2 πn + \frac{5 π}{6}

化简

2 x - \frac{5 π}{6} + \frac{5 π}{6} : 2 x

2 x - \frac{5 π}{6} + \frac{5 π}{6}

同类项相加:

- \frac{5 π}{6} + \frac{5 π}{6} = 0

= 2 x

化简

\frac{7 π}{6} + 2 πn + \frac{5 π}{6} : 2 πn + 2 π

\frac{7 π}{6} + 2 πn + \frac{5 π}{6}

对同类项分组

= 2 πn + \frac{7 π}{6} + \frac{5 π}{6}

合并分式

\frac{7 π}{6} + \frac{5 π}{6} : 2 π

使用法则

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

= \frac{7 π + 5 π}{6}

同类项相加:

7 π + 5 π = 12 π

= \frac{12 π}{6}

数字相除:

\frac{12}{6} = 2

= 2 π

= 2 πn + 2 π

2 x = 2 πn + 2 π

2 x = 2 πn + 2 π

2 x = 2 πn + 2 π

两边除以

2

2 x = 2 πn + 2 π

两边除以

2

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

化简

x = πn + π

x = πn + π

解

2 x - 5 \frac{π}{6} = \frac{11 π}{6} + 2 πn : x = πn + \frac{4 π}{3}

2 x - 5 \cdot \frac{π}{6} = \frac{11 π}{6} + 2 πn

乘

5 \cdot \frac{π}{6} : \frac{5 π}{6}

5 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 5}{6}

2 x - \frac{5 π}{6} = \frac{11 π}{6} + 2 πn

将

\frac{5 π}{6}

到右边

2 x - \frac{5 π}{6} = \frac{11 π}{6} + 2 πn

两边加上

\frac{5 π}{6}

2 x - \frac{5 π}{6} + \frac{5 π}{6} = \frac{11 π}{6} + 2 πn + \frac{5 π}{6}

化简

2 x - \frac{5 π}{6} + \frac{5 π}{6} = \frac{11 π}{6} + 2 πn + \frac{5 π}{6}

化简

2 x - \frac{5 π}{6} + \frac{5 π}{6} : 2 x

2 x - \frac{5 π}{6} + \frac{5 π}{6}

同类项相加:

- \frac{5 π}{6} + \frac{5 π}{6} = 0

= 2 x

化简

\frac{11 π}{6} + 2 πn + \frac{5 π}{6} : 2 πn + \frac{8 π}{3}

\frac{11 π}{6} + 2 πn + \frac{5 π}{6}

对同类项分组

= 2 πn + \frac{5 π}{6} + \frac{11 π}{6}

合并分式

\frac{5 π}{6} + \frac{11 π}{6} : \frac{8 π}{3}

使用法则

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

= \frac{5 π + 11 π}{6}

同类项相加:

5 π + 11 π = 16 π

= \frac{16 π}{6}

约分:

2

= \frac{8 π}{3}

= 2 πn + \frac{8 π}{3}

2 x = 2 πn + \frac{8 π}{3}

2 x = 2 πn + \frac{8 π}{3}

2 x = 2 πn + \frac{8 π}{3}

两边除以

2

2 x = 2 πn + \frac{8 π}{3}

两边除以

2

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

化简

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{8 π}{3}}{2} = \frac{4 π}{3}

\frac{\frac{8 π}{3}}{2}

使用分式法则:

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

= \frac{8 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{8 π}{6}

约分:

2

= \frac{4 π}{3}

= πn + \frac{4 π}{3}

x = πn + \frac{4 π}{3}

x = πn + \frac{4 π}{3}

x = πn + \frac{4 π}{3}

x = πn + π, x = πn + \frac{4 π}{3}

在

0 \leq x \leq 2 π

范围内的解

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

作图

流行的例子

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sin (x) = 1, 0 \leq x \leq 2 π

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solvefor x,f[g]=cos(1/(x^2))

so l v e f or x, f [g] = cos (\frac{1}{x ^{2}})

2cos(pi/5 θ)=sqrt(3)

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