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受欢迎的三角函数 >

cos(2x)> 1/(sqrt(2))

解答

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

解答

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

+2

间隔符号

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

十进制

- 0.39269 \dots + πn < x < 0.39269 \dots + πn

求解步骤

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

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}

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

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

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

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

合并区间

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

合并重叠的区间

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

流行的例子

sin(x)>= cos(x)

sin (x) \geq cos (x)

cos(x)<(sqrt(3))/2

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

2sin^2(x)-5sin(x)-3>= 0,0<= x<= 2pi

2 sin^{2} (x) - 5 sin (x) - 3 \geq 0, 0 \leq x \leq 2 π

tan(2x)<sqrt(3)

tan (2 x) < 3

2 cos (x) + 1 \leq 0