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受欢迎的三角函数 >

sin(pi/2 (x+1))= 2/5

解答

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

解答

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

+1

度数

x = - 42.28544 \dots^{\circ} + 229.18311 \dots^{\circ} n, x = 42.28544 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn, \frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn, \frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn

解

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn : x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn

在两边乘以

2

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn

在两边乘以

2

2 \cdot \frac{π}{2} (x + 1) = 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} (x + 1) = 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} (x + 1) : π (x + 1)

2 \cdot \frac{π}{2} (x + 1)

分式相乘:

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

= \frac{2 π}{2} (x + 1)

约分:

2

= (x + 1) π

化简

2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn : 2 arcsin (\frac{2}{5}) + 4 πn

2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

两边除以

π

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

两边除以

π

\frac{π ( x + 1 )}{π} = \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

化简

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

将

1

到右边

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

两边减去

1

x + 1 - 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

化简

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

解

\frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn : x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

\frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn

在两边乘以

2

\frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn

在两边乘以

2

2 \cdot \frac{π}{2} (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} (x + 1) : π (x + 1)

2 \cdot \frac{π}{2} (x + 1)

分式相乘:

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

= \frac{2 π}{2} (x + 1)

约分:

2

= (x + 1) π

化简

2 π - 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn : 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

2 π - 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

两边除以

π

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

两边除以

π

\frac{π ( x + 1 )}{π} = \frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

化简

\frac{π ( x + 1 )}{π} = \frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

化简

\frac{π ( x + 1 )}{π} : x + 1

\frac{π ( x + 1 )}{π}

约分:

π

= x + 1

化简

\frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π} : 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

\frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

消掉

\frac{2 π}{π} : 2

\frac{2 π}{π}

约分:

π

= 2

= 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

消掉

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

将

1

到右边

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

两边减去

1

x + 1 - 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

化简

x + 1 - 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

化简

x + 1 - 1 : x

x + 1 - 1

同类项相加:

1 - 1 = 0

= x

化简

2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1 : 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

数字相减:

2 - 1 = 1

= 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

以小数形式表示解

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

作图

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