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受欢迎的三角函数 >

3cos(2x)+2< 5/2

解答

3 cos (2 x) + 2 < \frac{5}{2}

解答

\frac{arccos ( \frac{1}{6} )}{2} + πn < x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

+2

间隔符号

(\frac{arccos ( \frac{1}{6} )}{2} + πn, \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn)

十进制

0.70167 \dots + πn < x < 2.43991 \dots + πn

求解步骤

3 cos (2 x) + 2 < \frac{5}{2}

将

2

到右边

3 cos (2 x) + 2 < \frac{5}{2}

两边减去

2

3 cos (2 x) + 2 - 2 < \frac{5}{2} - 2

化简

3 cos (2 x) + 2 - 2 < \frac{5}{2} - 2

化简

3 cos (2 x) + 2 - 2 : 3 cos (2 x)

3 cos (2 x) + 2 - 2

同类项相加:

2 - 2 < 0

= 3 cos (2 x)

化简

\frac{5}{2} - 2 : \frac{1}{2}

\frac{5}{2} - 2

将项转换为分式:

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

= - \frac{2 \cdot 2}{2} + \frac{5}{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 + 5}{2}

- 2 \cdot 2 + 5 = 1

- 2 \cdot 2 + 5

数字相乘:

2 \cdot 2 = 4

= - 4 + 5

数字相加/相减:

- 4 + 5 = 1

= 1

= \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

3 cos (2 x) < \frac{1}{2}

两边除以

3

3 cos (2 x) < \frac{1}{2}

两边除以

3

\frac{3 cos ( 2 x )}{3} < \frac{\frac{1}{2}}{3}

化简

\frac{3 cos ( 2 x )}{3} < \frac{\frac{1}{2}}{3}

化简

\frac{3 cos ( 2 x )}{3} : cos (2 x)

\frac{3 cos ( 2 x )}{3}

数字相除:

\frac{3}{3} = 1

= cos (2 x)

化简

\frac{\frac{1}{2}}{3} : \frac{1}{6}

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

使用分式法则:

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

= \frac{1}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{1}{6}

cos (2 x) < \frac{1}{6}

cos (2 x) < \frac{1}{6}

cos (2 x) < \frac{1}{6}

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

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

arccos (\frac{1}{6}) + 2 πn < 2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

若

a < u < b

，则

a < u and u < b

arccos (\frac{1}{6}) + 2 πn < 2 x and 2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

arccos (\frac{1}{6}) + 2 πn < 2 x : x > \frac{arccos ( \frac{1}{6} )}{2} + πn

arccos (\frac{1}{6}) + 2 πn < 2 x

交换两边

2 x > arccos (\frac{1}{6}) + 2 πn

两边除以

2

2 x > arccos (\frac{1}{6}) + 2 πn

两边除以

2

\frac{2 x}{2} > \frac{arccos ( \frac{1}{6} )}{2} + \frac{2 πn}{2}

化简

x > \frac{arccos ( \frac{1}{6} )}{2} + πn

x > \frac{arccos ( \frac{1}{6} )}{2} + πn

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn : x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

两边除以

2

2 x < 2 π - arccos (\frac{1}{6}) + 2 πn

两边除以

2

\frac{2 x}{2} < \frac{2 π}{2} - \frac{arccos ( \frac{1}{6} )}{2} + \frac{2 πn}{2}

化简

x < π - \frac{arccos ( \frac{1}{6} )}{2} + πn

化简

π - \frac{arccos ( \frac{1}{6} )}{2} : \frac{2 π - arccos ( \frac{1}{6} )}{2}

π - \frac{arccos ( \frac{1}{6} )}{2}

将项转换为分式:

π = \frac{π 2}{2}

= \frac{π 2}{2} - \frac{arccos ( \frac{1}{6} )}{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{π 2 - arccos ( \frac{1}{6} )}{2}

x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

合并区间

x > \frac{arccos ( \frac{1}{6} )}{2} + πn and x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

合并重叠的区间

\frac{arccos ( \frac{1}{6} )}{2} + πn < x < \frac{2 π - arccos ( \frac{1}{6} )}{2} + πn

流行的例子

sin(x)-cos(x)<0

sin (x) - cos (x) < 0

cos (3 x) \geq 9

5 sin (x) \leq 3

- sin (x) \leq \frac{2}{π}

sin(x)>-1,0<= x<= 2pi

sin (x) > - 1, 0 \leq x \leq 2 π