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受欢迎的三角函数 >

0<2cos(2x)-sqrt(3)

解答

0 < 2 cos (2 x) - 3

解答

- \frac{π}{12} + πn < x < \frac{π}{12} + πn

+2

间隔符号

(- \frac{π}{12} + πn, \frac{π}{12} + πn)

十进制

- 0.26179 \dots + πn < x < 0.26179 \dots + πn

求解步骤

0 < 2 cos (2 x) - 3

交换两边

2 cos (2 x) - 3 > 0

将

3

到右边

2 cos (2 x) - 3 > 0

两边加上

3

2 cos (2 x) - 3 + 3 > 0 + 3

化简

2 cos (2 x) > 3

2 cos (2 x) > 3

两边除以

2

2 cos (2 x) > 3

两边除以

2

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

化简

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

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

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

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

- arccos (\frac{3}{2}) + 2 πn < 2 x < arccos (\frac{3}{2}) + 2 πn

若

a < u < b

，则

a < u and u < b

- arccos (\frac{3}{2}) + 2 πn < 2 x and 2 x < arccos (\frac{3}{2}) + 2 πn

- arccos (\frac{3}{2}) + 2 πn < 2 x : x > - \frac{π}{12} + πn

- arccos (\frac{3}{2}) + 2 πn < 2 x

交换两边

2 x > - arccos (\frac{3}{2}) + 2 πn

化简

- arccos (\frac{3}{2}) + 2 πn : - \frac{π}{6} + 2 πn

- arccos (\frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{6} + 2 πn

2 x > - \frac{π}{6} + 2 πn

两边除以

2

2 x > - \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} > - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} > - \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- \frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : - \frac{π}{12} + πn

- \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

x > - \frac{π}{12} + πn

2 x < arccos (\frac{3}{2}) + 2 πn : x < \frac{π}{12} + πn

2 x < arccos (\frac{3}{2}) + 2 πn

化简

arccos (\frac{3}{2}) + 2 πn : \frac{π}{6} + 2 πn

arccos (\frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6} + 2 πn

2 x < \frac{π}{6} + 2 πn

两边除以

2

2 x < \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} < \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} < \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x < \frac{π}{12} + πn

x < \frac{π}{12} + πn

x < \frac{π}{12} + πn

合并区间

x > - \frac{π}{12} + πn and x < \frac{π}{12} + πn

合并重叠的区间

- \frac{π}{12} + πn < x < \frac{π}{12} + πn

流行的例子

1 - sin (x) \geq 0

cot (θ) > 0

sin(x)<= (sqrt(3))/2

sin (x) \leq \frac{3}{2}

2cos^2(x)>= 1.5

2 cos^{2} (x) \geq 1.5

sin(x)>= (sqrt(2))/2

sin (x) \geq \frac{2}{2}