zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

sqrt(2)cos(2x)<1

解答

2 cos (2 x) < 1

解答

\frac{π}{8} + πn < x < \frac{7 π}{8} + πn

+2

间隔符号

(\frac{π}{8} + πn, \frac{7 π}{8} + πn)

十进制

0.39269 \dots + πn < x < 2.74889 \dots + πn

求解步骤

2 cos (2 x) < 1

两边除以

2

2 cos (2 x) < 1

两边除以

2

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

化简

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

化简

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

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

约分:

2

= cos (2 x)

化简

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

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

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

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

若

a < u < b

，则

a < u and u < b

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

arccos (\frac{2}{2}) + 2 πn < 2 x : x > \frac{π}{8} + πn

arccos (\frac{2}{2}) + 2 πn < 2 x

交换两边

2 x > arccos (\frac{2}{2}) + 2 πn

化简

arccos (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

2 x > \frac{π}{4} + 2 πn

两边除以

2

2 x > \frac{π}{4} + 2 πn

两边除以

2

\frac{2 x}{2} > \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} > \frac{\frac{π}{4}}{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}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x > \frac{π}{8} + πn

x > \frac{π}{8} + πn

x > \frac{π}{8} + πn

2 x < 2 π - arccos (\frac{2}{2}) + 2 πn : x < \frac{7 π}{8} + πn

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

化简

2 π - arccos (\frac{2}{2}) + 2 πn : 2 π - \frac{π}{4} + 2 πn

2 π - arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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}

= 2 π - \frac{π}{4} + 2 πn

2 x < 2 π - \frac{π}{4} + 2 πn

两边除以

2

2 x < 2 π - \frac{π}{4} + 2 πn

两边除以

2

\frac{2 x}{2} < \frac{2 π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} < \frac{2 π}{2} - \frac{\frac{π}{4}}{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{2 π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : π - \frac{π}{8} + πn

\frac{2 π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= π - \frac{π}{8} + πn

x < π - \frac{π}{8} + πn

x < π - \frac{π}{8} + πn

化简

π - \frac{π}{8} : \frac{7 π}{8}

π - \frac{π}{8}

将项转换为分式:

π = \frac{π 8}{8}

= \frac{π 8}{8} - \frac{π}{8}

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 8 - π}{8}

同类项相加:

8 π - π = 7 π

= \frac{7 π}{8}

x < \frac{7 π}{8} + πn

x < \frac{7 π}{8} + πn

合并区间

x > \frac{π}{8} + πn and x < \frac{7 π}{8} + πn

合并重叠的区间

\frac{π}{8} + πn < x < \frac{7 π}{8} + πn

流行的例子

12 cos (2 x) < 0

0.86<= cos^{2(9)}((68)/n)

0.86 \leq cos^{2 (9)} (\frac{68}{n})

cos(x)<= 1+sin(x)

cos (x) \leq 1 + sin (x)

cos(θ)<0,csc(θ)<0

cos (θ) < 0, csc (θ) < 0

50sin(-(2pi)/3 x-pi/2)>=-15

50 sin (- \frac{2 π}{3} x - \frac{π}{2}) \geq - 15