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受欢迎的三角函数 >

10<5sin(pi/6 (x+2))+6

解答

10 < 5 sin (\frac{π}{6} (x + 2)) + 6

解答

\frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n < x < \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n

+2

间隔符号

(\frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n, \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n)

十进制

- 0.22899 \dots + 12 n < x < 2.22899 \dots + 12 n

求解步骤

10 < 5 sin (\frac{π}{6} (x + 2)) + 6

交换两边

5 sin (\frac{π}{6} (x + 2)) + 6 > 10

将

6

到右边

5 sin (\frac{π}{6} (x + 2)) + 6 > 10

两边减去

6

5 sin (\frac{π}{6} (x + 2)) + 6 - 6 > 10 - 6

化简

5 sin (\frac{π}{6} (x + 2)) > 4

5 sin (\frac{π}{6} (x + 2)) > 4

两边除以

5

5 sin (\frac{π}{6} (x + 2)) > 4

两边除以

5

\frac{5 sin ( \frac{π}{6} ( x + 2 ) )}{5} > \frac{4}{5}

化简

sin (\frac{π}{6} (x + 2)) > \frac{4}{5}

sin (\frac{π}{6} (x + 2)) > \frac{4}{5}

对于

sin (x) > a

，若

- 1 \leq a < 1

，则

arcsin (a) + 2 πn < x < π - arcsin (a) + 2 πn

arcsin (\frac{4}{5}) + 2 πn < \frac{π}{6} (x + 2) < π - arcsin (\frac{4}{5}) + 2 πn

若

a < u < b

，则

a < u and u < b

arcsin (\frac{4}{5}) + 2 πn < \frac{π}{6} (x + 2) and \frac{π}{6} (x + 2) < π - arcsin (\frac{4}{5}) + 2 πn

arcsin (\frac{4}{5}) + 2 πn < \frac{π}{6} (x + 2) : x > \frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n

arcsin (\frac{4}{5}) + 2 πn < \frac{π}{6} (x + 2)

交换两边

\frac{π}{6} (x + 2) > arcsin (\frac{4}{5}) + 2 πn

在两边乘以

6

\frac{π}{6} (x + 2) > arcsin (\frac{4}{5}) + 2 πn

在两边乘以

6

6 \cdot \frac{π}{6} (x + 2) > 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

化简

6 \cdot \frac{π}{6} (x + 2) > 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

化简

6 \cdot \frac{π}{6} (x + 2) : π (x + 2)

6 \cdot \frac{π}{6} (x + 2)

分式相乘:

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

= \frac{6 π}{6} (x + 2)

约分:

6

= (x + 2) π

化简

6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn : 6 arcsin (\frac{4}{5}) + 12 πn

6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

数字相乘:

6 \cdot 2 = 12

= 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) > 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) > 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) > 6 arcsin (\frac{4}{5}) + 12 πn

两边除以

π

π (x + 2) > 6 arcsin (\frac{4}{5}) + 12 πn

两边除以

π

\frac{π ( x + 2 )}{π} > \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π}

化简

x + 2 > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

x + 2 > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

将

2

到右边

x + 2 > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

两边减去

2

x + 2 - 2 > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

化简

x > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

x > \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

化简

\frac{6 arcsin ( \frac{4}{5} )}{π} - 2 : \frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π}

\frac{6 arcsin ( \frac{4}{5} )}{π} - 2

将项转换为分式:

2 = \frac{2 π}{π}

= \frac{6 arcsin ( \frac{4}{5} )}{π} - \frac{2 π}{π}

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{6 arcsin ( \frac{4}{5} ) - 2 π}{π}

x > \frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n

\frac{π}{6} (x + 2) < π - arcsin (\frac{4}{5}) + 2 πn : x < \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n

\frac{π}{6} (x + 2) < π - arcsin (\frac{4}{5}) + 2 πn

在两边乘以

6

\frac{π}{6} (x + 2) < π - arcsin (\frac{4}{5}) + 2 πn

在两边乘以

6

6 \cdot \frac{π}{6} (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

化简

6 \cdot \frac{π}{6} (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

化简

6 \cdot \frac{π}{6} (x + 2) : π (x + 2)

6 \cdot \frac{π}{6} (x + 2)

分式相乘:

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

= \frac{6 π}{6} (x + 2)

约分:

6

= (x + 2) π

化简

6 π - 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn : 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

6 π - 6 arcsin (\frac{4}{5}) + 6 \cdot 2 πn

数字相乘:

6 \cdot 2 = 12

= 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

π (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

两边除以

π

π (x + 2) < 6 π - 6 arcsin (\frac{4}{5}) + 12 πn

两边除以

π

\frac{π ( x + 2 )}{π} < \frac{6 π}{π} - \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π}

化简

\frac{π ( x + 2 )}{π} < \frac{6 π}{π} - \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π}

化简

\frac{π ( x + 2 )}{π} : x + 2

\frac{π ( x + 2 )}{π}

约分:

π

= x + 2

化简

\frac{6 π}{π} - \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π} : 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

\frac{6 π}{π} - \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π}

消掉

\frac{6 π}{π} : 6

\frac{6 π}{π}

约分:

π

= 6

= 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + \frac{12 πn}{π}

消掉

\frac{12 πn}{π} : 12 n

\frac{12 πn}{π}

约分:

π

= 12 n

= 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

x + 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

x + 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

x + 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

将

2

到右边

x + 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n

两边减去

2

x + 2 - 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

化简

x + 2 - 2 < 6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

化简

x + 2 - 2 : x

x + 2 - 2

同类项相加:

2 - 2 < 0

= x

化简

6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2 : 12 n + 4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

6 - \frac{6 arcsin ( \frac{4}{5} )}{π} + 12 n - 2

数字相减:

6 - 2 = 4

= 12 n + 4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

x < 12 n + 4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

x < 12 n + 4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

x < 12 n + 4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

化简

4 - \frac{6 arcsin ( \frac{4}{5} )}{π} : \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π}

4 - \frac{6 arcsin ( \frac{4}{5} )}{π}

将项转换为分式:

4 = \frac{4 π}{π}

= \frac{4 π}{π} - \frac{6 arcsin ( \frac{4}{5} )}{π}

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{4 π - 6 arcsin ( \frac{4}{5} )}{π}

x < \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n

合并区间

x > \frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n and x < \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n

合并重叠的区间

\frac{6 arcsin ( \frac{4}{5} ) - 2 π}{π} + 12 n < x < \frac{4 π - 6 arcsin ( \frac{4}{5} )}{π} + 12 n

流行的例子

arctan(1-|x|)<= 1/2

arctan (1 - ∣ x ∣) \leq \frac{1}{2}

sin(t)< 1/3 ln(1)

sin (t) < \frac{1}{3} ln (1)

5-8cos(x)+4cos(2x)<0

5 - 8 cos (x) + 4 cos (2 x) < 0

sin (a) < 0

sin^{(2)}(x)<= ((1))/((2))

sin^{(2)} (x) \leq \frac{( 1 )}{( 2 )}