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受欢迎的三角函数 >

0=4sin(pi/2 (x-2))+2

解答

0 = 4 sin (\frac{π}{2} (x - 2)) + 2

解答

x = 4 n + \frac{13}{3}, x = 4 n + \frac{17}{3}

+1

度数

x = 248.28171 \dots^{\circ} + 229.18311 \dots^{\circ} n, x = 324.67608 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

0 = 4 sin (\frac{π}{2} (x - 2)) + 2

交换两边

4 sin (\frac{π}{2} (x - 2)) + 2 = 0

将

2

到右边

4 sin (\frac{π}{2} (x - 2)) + 2 = 0

两边减去

2

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

化简

4 sin (\frac{π}{2} (x - 2)) = - 2

4 sin (\frac{π}{2} (x - 2)) = - 2

两边除以

4

4 sin (\frac{π}{2} (x - 2)) = - 2

两边除以

4

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

化简

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

化简

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

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

数字相除:

\frac{4}{4} = 1

= sin (\frac{π}{2} (x - 2))

化简

\frac{- 2}{4} : - \frac{1}{2}

\frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

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

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

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

sin (\frac{π}{2} (x - 2)) = - \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}

\frac{π}{2} (x - 2) = \frac{7 π}{6} + 2 πn, \frac{π}{2} (x - 2) = \frac{11 π}{6} + 2 πn

\frac{π}{2} (x - 2) = \frac{7 π}{6} + 2 πn, \frac{π}{2} (x - 2) = \frac{11 π}{6} + 2 πn

解

\frac{π}{2} (x - 2) = \frac{7 π}{6} + 2 πn : x = 4 n + \frac{13}{3}

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

在两边乘以

2

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

在两边乘以

2

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

化简

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

化简

2 \cdot \frac{π}{2} (x - 2) : π (x - 2)

2 \cdot \frac{π}{2} (x - 2)

分式相乘:

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

= \frac{2 π}{2} (x - 2)

约分:

2

= (x - 2) π

化简

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn : \frac{7 π}{3} + 4 πn

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{6} = \frac{7 π}{3}

2 \cdot \frac{7 π}{6}

分式相乘:

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

= \frac{7 π 2}{6}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{6}

约分:

2

= \frac{7 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{3} + 4 πn

π (x - 2) = \frac{7 π}{3} + 4 πn

π (x - 2) = \frac{7 π}{3} + 4 πn

π (x - 2) = \frac{7 π}{3} + 4 πn

两边除以

π

π (x - 2) = \frac{7 π}{3} + 4 πn

两边除以

π

\frac{π ( x - 2 )}{π} = \frac{\frac{7 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π ( x - 2 )}{π} = \frac{\frac{7 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π ( x - 2 )}{π} : x - 2

\frac{π ( x - 2 )}{π}

约分:

π

= x - 2

化简

\frac{\frac{7 π}{3}}{π} + \frac{4 πn}{π} : \frac{7}{3} + 4 n

\frac{\frac{7 π}{3}}{π} + \frac{4 πn}{π}

\frac{\frac{7 π}{3}}{π} = \frac{7}{3}

\frac{\frac{7 π}{3}}{π}

使用分式法则:

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

= \frac{7 π}{3 π}

约分:

π

= \frac{7}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{7}{3} + 4 n

x - 2 = \frac{7}{3} + 4 n

x - 2 = \frac{7}{3} + 4 n

x - 2 = \frac{7}{3} + 4 n

将

2

到右边

x - 2 = \frac{7}{3} + 4 n

两边加上

2

x - 2 + 2 = \frac{7}{3} + 4 n + 2

化简

x - 2 + 2 = \frac{7}{3} + 4 n + 2

化简

x - 2 + 2 : x

x - 2 + 2

同类项相加:

- 2 + 2 = 0

= x

化简

\frac{7}{3} + 4 n + 2 : 4 n + \frac{13}{3}

\frac{7}{3} + 4 n + 2

合并分式

2 + \frac{7}{3} : \frac{13}{3}

2 + \frac{7}{3}

将项转换为分式:

2 = \frac{2 \cdot 3}{3}

= \frac{2 \cdot 3}{3} + \frac{7}{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 \cdot 3 + 7}{3}

2 \cdot 3 + 7 = 13

2 \cdot 3 + 7

数字相乘:

2 \cdot 3 = 6

= 6 + 7

数字相加:

6 + 7 = 13

= 13

= \frac{13}{3}

= 4 n + \frac{13}{3}

x = 4 n + \frac{13}{3}

x = 4 n + \frac{13}{3}

x = 4 n + \frac{13}{3}

解

\frac{π}{2} (x - 2) = \frac{11 π}{6} + 2 πn : x = 4 n + \frac{17}{3}

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

在两边乘以

2

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

在两边乘以

2

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

化简

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

化简

2 \cdot \frac{π}{2} (x - 2) : π (x - 2)

2 \cdot \frac{π}{2} (x - 2)

分式相乘:

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

= \frac{2 π}{2} (x - 2)

约分:

2

= (x - 2) π

化简

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

分式相乘:

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

= \frac{11 π 2}{6}

数字相乘:

11 \cdot 2 = 22

= \frac{22 π}{6}

约分:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

π (x - 2) = \frac{11 π}{3} + 4 πn

π (x - 2) = \frac{11 π}{3} + 4 πn

π (x - 2) = \frac{11 π}{3} + 4 πn

两边除以

π

π (x - 2) = \frac{11 π}{3} + 4 πn

两边除以

π

\frac{π ( x - 2 )}{π} = \frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π ( x - 2 )}{π} = \frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π ( x - 2 )}{π} : x - 2

\frac{π ( x - 2 )}{π}

约分:

π

= x - 2

化简

\frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π} : \frac{11}{3} + 4 n

\frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

\frac{\frac{11 π}{3}}{π} = \frac{11}{3}

\frac{\frac{11 π}{3}}{π}

使用分式法则:

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

= \frac{11 π}{3 π}

约分:

π

= \frac{11}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{11}{3} + 4 n

x - 2 = \frac{11}{3} + 4 n

x - 2 = \frac{11}{3} + 4 n

x - 2 = \frac{11}{3} + 4 n

将

2

到右边

x - 2 = \frac{11}{3} + 4 n

两边加上

2

x - 2 + 2 = \frac{11}{3} + 4 n + 2

化简

x - 2 + 2 = \frac{11}{3} + 4 n + 2

化简

x - 2 + 2 : x

x - 2 + 2

同类项相加:

- 2 + 2 = 0

= x

化简

\frac{11}{3} + 4 n + 2 : 4 n + \frac{17}{3}

\frac{11}{3} + 4 n + 2

合并分式

2 + \frac{11}{3} : \frac{17}{3}

2 + \frac{11}{3}

将项转换为分式:

2 = \frac{2 \cdot 3}{3}

= \frac{2 \cdot 3}{3} + \frac{11}{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 \cdot 3 + 11}{3}

2 \cdot 3 + 11 = 17

2 \cdot 3 + 11

数字相乘:

2 \cdot 3 = 6

= 6 + 11

数字相加:

6 + 11 = 17

= 17

= \frac{17}{3}

= 4 n + \frac{17}{3}

x = 4 n + \frac{17}{3}

x = 4 n + \frac{17}{3}

x = 4 n + \frac{17}{3}

x = 4 n + \frac{13}{3}, x = 4 n + \frac{17}{3}

作图

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