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受欢迎的三角函数 >

7/2 =3sin(b(pi/2))+2

解答

\frac{7}{2} = 3 sin (b (\frac{π}{2})) + 2

解答

b = \frac{1}{3} + 4 n, b = \frac{5}{3} + 4 n

+1

度数

b = 19.09859 \dots^{\circ} + 229.18311 \dots^{\circ} n, b = 95.49296 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

\frac{7}{2} = 3 sin (b (\frac{π}{2})) + 2

交换两边

3 sin (b \frac{π}{2}) + 2 = \frac{7}{2}

将

2

到右边

3 sin (b \frac{π}{2}) + 2 = \frac{7}{2}

两边减去

2

3 sin (b \frac{π}{2}) + 2 - 2 = \frac{7}{2} - 2

化简

3 sin (b \frac{π}{2}) + 2 - 2 = \frac{7}{2} - 2

化简

3 sin (b \frac{π}{2}) + 2 - 2 : 3 sin (b \frac{π}{2})

3 sin (b \frac{π}{2}) + 2 - 2

同类项相加:

2 - 2 = 0

= 3 sin (b \frac{π}{2})

化简

\frac{7}{2} - 2 : \frac{3}{2}

\frac{7}{2} - 2

将项转换为分式:

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

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

- 2 \cdot 2 + 7 = 3

- 2 \cdot 2 + 7

数字相乘:

2 \cdot 2 = 4

= - 4 + 7

数字相加/相减:

- 4 + 7 = 3

= 3

= \frac{3}{2}

3 sin (b \frac{π}{2}) = \frac{3}{2}

3 sin (b \frac{π}{2}) = \frac{3}{2}

3 sin (b \frac{π}{2}) = \frac{3}{2}

两边除以

3

3 sin (b \frac{π}{2}) = \frac{3}{2}

两边除以

3

\frac{3 sin ( b \frac{π}{2} )}{3} = \frac{\frac{3}{2}}{3}

化简

\frac{3 sin ( b \frac{π}{2} )}{3} = \frac{\frac{3}{2}}{3}

化简

\frac{3 sin ( b \frac{π}{2} )}{3} : sin (b \frac{π}{2})

\frac{3 sin ( b \frac{π}{2} )}{3}

数字相除:

\frac{3}{3} = 1

= sin (b \frac{π}{2})

化简

\frac{\frac{3}{2}}{3} : \frac{1}{2}

\frac{\frac{3}{2}}{3}

使用分式法则:

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

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

数字相乘:

2 \cdot 3 = 6

= \frac{3}{6}

约分:

3

= \frac{1}{2}

sin (b \frac{π}{2}) = \frac{1}{2}

sin (b \frac{π}{2}) = \frac{1}{2}

sin (b \frac{π}{2}) = \frac{1}{2}

sin (b \frac{π}{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}

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

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

解

b \frac{π}{2} = \frac{π}{6} + 2 πn : b = \frac{1}{3} + 4 n

b \frac{π}{2} = \frac{π}{6} + 2 πn

在两边乘以

2

b \frac{π}{2} = \frac{π}{6} + 2 πn

在两边乘以

2

2 b \frac{π}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

2 b \frac{π}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

2 b \frac{π}{2} : πb

2 b \frac{π}{2}

分式相乘:

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

= \frac{2 π}{2} b

约分:

2

= bπ

化简

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

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 2}{6}

约分:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

πb = \frac{π}{3} + 4 πn

πb = \frac{π}{3} + 4 πn

πb = \frac{π}{3} + 4 πn

两边除以

π

πb = \frac{π}{3} + 4 πn

两边除以

π

\frac{πb}{π} = \frac{\frac{π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{πb}{π} = \frac{\frac{π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{πb}{π} : b

\frac{πb}{π}

约分:

π

= b

化简

\frac{\frac{π}{3}}{π} + \frac{4 πn}{π} : \frac{1}{3} + 4 n

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

\frac{\frac{π}{3}}{π} = \frac{1}{3}

\frac{\frac{π}{3}}{π}

使用分式法则:

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

= \frac{π}{3 π}

约分:

π

= \frac{1}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{1}{3} + 4 n

b = \frac{1}{3} + 4 n

b = \frac{1}{3} + 4 n

b = \frac{1}{3} + 4 n

解

b \frac{π}{2} = \frac{5 π}{6} + 2 πn : b = \frac{5}{3} + 4 n

b \frac{π}{2} = \frac{5 π}{6} + 2 πn

在两边乘以

2

b \frac{π}{2} = \frac{5 π}{6} + 2 πn

在两边乘以

2

2 b \frac{π}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

化简

2 b \frac{π}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

化简

2 b \frac{π}{2} : πb

2 b \frac{π}{2}

分式相乘:

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

= \frac{2 π}{2} b

约分:

2

= bπ

化简

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

分式相乘:

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

= \frac{5 π 2}{6}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{6}

约分:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

πb = \frac{5 π}{3} + 4 πn

πb = \frac{5 π}{3} + 4 πn

πb = \frac{5 π}{3} + 4 πn

两边除以

π

πb = \frac{5 π}{3} + 4 πn

两边除以

π

\frac{πb}{π} = \frac{\frac{5 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{πb}{π} = \frac{\frac{5 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{πb}{π} : b

\frac{πb}{π}

约分:

π

= b

化简

\frac{\frac{5 π}{3}}{π} + \frac{4 πn}{π} : \frac{5}{3} + 4 n

\frac{\frac{5 π}{3}}{π} + \frac{4 πn}{π}

\frac{\frac{5 π}{3}}{π} = \frac{5}{3}

\frac{\frac{5 π}{3}}{π}

使用分式法则:

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

= \frac{5 π}{3 π}

约分:

π

= \frac{5}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{5}{3} + 4 n

b = \frac{5}{3} + 4 n

b = \frac{5}{3} + 4 n

b = \frac{5}{3} + 4 n

b = \frac{1}{3} + 4 n, b = \frac{5}{3} + 4 n

作图

流行的例子

cos (x) = \frac{2}{8}

solvefor y,x=4sin(2y)

so l v e f or y, x = 4 sin (2 y)

cos^2(x)-5/2 cos(x)-3/2 =0

cos^{2} (x) - \frac{5}{2} cos (x) - \frac{3}{2} = 0

solvefor x,sin(4x)= 1/2

so l v e f or x, sin (4 x) = \frac{1}{2}

sec(2x-10)=csc(50)

sec (2 x - 1 0^{\circ}) = csc (5 0^{\circ})