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受欢迎的三角函数 >

sin(2x+20)-1=(-1)/2

解答

sin (2 x + 20) - 1 = \frac{- 1}{2}

解答

x = - 10 + πn + \frac{π}{12}, x = - 10 + πn + \frac{5 π}{12}

+1

度数

x = - 557.95779 \dots^{\circ} + 18 0^{\circ} n, x = - 497.95779 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sin (2 x + 20) - 1 = \frac{- 1}{2}

将

1

到右边

sin (2 x + 20) - 1 = \frac{- 1}{2}

两边加上

1

sin (2 x + 20) - 1 + 1 = \frac{- 1}{2} + 1

化简

sin (2 x + 20) - 1 + 1 = \frac{- 1}{2} + 1

化简

sin (2 x + 20) - 1 + 1 : sin (2 x + 20)

sin (2 x + 20) - 1 + 1

同类项相加:

- 1 + 1 = 0

= sin (2 x + 20)

化简

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

\frac{- 1}{2} + 1

使用分式法则:

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

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

将项转换为分式:

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

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

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

数字相乘:

1 \cdot 2 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

sin (2 x + 20) = \frac{1}{2}

sin (2 x + 20) = \frac{1}{2}

sin (2 x + 20) = \frac{1}{2}

sin (2 x + 20) = \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 + 20 = \frac{π}{6} + 2 πn, 2 x + 20 = \frac{5 π}{6} + 2 πn

2 x + 20 = \frac{π}{6} + 2 πn, 2 x + 20 = \frac{5 π}{6} + 2 πn

解

2 x + 20 = \frac{π}{6} + 2 πn : x = - 10 + πn + \frac{π}{12}

2 x + 20 = \frac{π}{6} + 2 πn

将

20

到右边

2 x + 20 = \frac{π}{6} + 2 πn

两边减去

20

2 x + 20 - 20 = \frac{π}{6} + 2 πn - 20

化简

2 x = \frac{π}{6} + 2 πn - 20

2 x = \frac{π}{6} + 2 πn - 20

两边除以

2

2 x = \frac{π}{6} + 2 πn - 20

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{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}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2} : - 10 + πn + \frac{π}{12}

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2}

对同类项分组

= - \frac{20}{2} + \frac{2 πn}{2} + \frac{\frac{π}{6}}{2}

\frac{20}{2} = 10

\frac{20}{2}

数字相除:

\frac{20}{2} = 10

= 10

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

= - 10 + πn + \frac{π}{12}

x = - 10 + πn + \frac{π}{12}

x = - 10 + πn + \frac{π}{12}

x = - 10 + πn + \frac{π}{12}

解

2 x + 20 = \frac{5 π}{6} + 2 πn : x = - 10 + πn + \frac{5 π}{12}

2 x + 20 = \frac{5 π}{6} + 2 πn

将

20

到右边

2 x + 20 = \frac{5 π}{6} + 2 πn

两边减去

20

2 x + 20 - 20 = \frac{5 π}{6} + 2 πn - 20

化简

2 x = \frac{5 π}{6} + 2 πn - 20

2 x = \frac{5 π}{6} + 2 πn - 20

两边除以

2

2 x = \frac{5 π}{6} + 2 πn - 20

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{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}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2} : - 10 + πn + \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} - \frac{20}{2}

对同类项分组

= - \frac{20}{2} + \frac{2 πn}{2} + \frac{\frac{5 π}{6}}{2}

\frac{20}{2} = 10

\frac{20}{2}

数字相除:

\frac{20}{2} = 10

= 10

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

= - 10 + πn + \frac{5 π}{12}

x = - 10 + πn + \frac{5 π}{12}

x = - 10 + πn + \frac{5 π}{12}

x = - 10 + πn + \frac{5 π}{12}

x = - 10 + πn + \frac{π}{12}, x = - 10 + πn + \frac{5 π}{12}

作图

流行的例子

solvefor x,sin(x+pi/2)-cos^2(x)=0

so l v e f or x, sin (x + \frac{π}{2}) - cos^{2} (x) = 0

0 = 4 + 3 cos (θ)

cos (θ) = \frac{6}{8}

solvefor x,4cos(x)*sin(y)=1

so l v e f or x, 4 cos (x) \cdot sin (y) = 1

arctan(1/4 x)-pi-arctan(x/(100))=-pi

arctan (\frac{1}{4} x) - π - arctan (\frac{x}{100}) = - π