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受欢迎的三角函数 >

2sin(x-pi/4)+1=2

解答

2 sin (x - \frac{π}{4}) + 1 = 2

解答

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

+1

度数

x = 7 5^{\circ} + 36 0^{\circ} n, x = 19 5^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x - \frac{π}{4}) + 1 = 2

将

1

到右边

2 sin (x - \frac{π}{4}) + 1 = 2

两边减去

1

2 sin (x - \frac{π}{4}) + 1 - 1 = 2 - 1

化简

2 sin (x - \frac{π}{4}) = 1

2 sin (x - \frac{π}{4}) = 1

两边除以

2

2 sin (x - \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{1}{2}

化简

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \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}

x - \frac{π}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

x - \frac{π}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

解

x - \frac{π}{4} = \frac{π}{6} + 2 πn : x = 2 πn + \frac{5 π}{12}

x - \frac{π}{4} = \frac{π}{6} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{π}{6} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{5 π}{12}

\frac{π}{6} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{6} + \frac{π}{4}

6, 4

的最小公倍数:

12

6, 4

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

6

或

4

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{6} :

将分母和分子乘以

2

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

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 2}{12} + \frac{π 3}{12}

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{π 2 + π 3}{12}

同类项相加:

2 π + 3 π = 5 π

= 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

解

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn : x = 2 πn + \frac{13 π}{12}

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{5 π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{13 π}{12}

\frac{5 π}{6} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{5 π}{6}

4, 6

的最小公倍数:

12

4, 6

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

4

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{5 π}{6} :

将分母和分子乘以

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= \frac{π 3}{12} + \frac{10 π}{12}

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{π 3 + 10 π}{12}

同类项相加:

3 π + 10 π = 13 π

= 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

作图

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