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受欢迎的三角函数 >

4cos(2x)>= 2,0<= x<2pi

解答

4 cos (2 x) \geq 2, 0 \leq x < 2 π

解答

0 \leq x \leq \frac{π}{6} or π - \frac{π}{6} \leq x \leq π + \frac{π}{6} or 2 π - \frac{π}{6} \leq x < 2 π

+2

间隔符号

[0, \frac{π}{6}] \cup [π - \frac{π}{6}, π + \frac{π}{6}] \cup [2 π - \frac{π}{6}, 2 π)

十进制

0 \leq x \leq 0.52359 \dots or 2.61799 \dots \leq x \leq 3.66519 \dots or 5.75958 \dots \leq x < 6.28318 \dots

求解步骤

4 cos (2 x) \geq 2, 0 \leq x < 2 π

两边除以

4

4 cos (2 x) \geq 2

两边除以

4

\frac{4 cos ( 2 x )}{4} \geq \frac{2}{4}

化简

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

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

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{1}{2}) + 2 πn \leq 2 x \leq arccos (\frac{1}{2}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (\frac{1}{2}) + 2 πn \leq 2 x and 2 x \leq arccos (\frac{1}{2}) + 2 πn

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

- arccos (\frac{1}{2}) + 2 πn \leq 2 x

交换两边

2 x \geq - arccos (\frac{1}{2}) + 2 πn

化简

- arccos (\frac{1}{2}) + 2 πn : - \frac{π}{3} + 2 πn

- arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

2 x \geq - \frac{π}{3} + 2 πn

两边除以

2

2 x \geq - \frac{π}{3} + 2 πn

两边除以

2

\frac{2 x}{2} \geq - \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \geq - \frac{\frac{π}{3}}{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}{3}}{2} + \frac{2 πn}{2} : - \frac{π}{6} + πn

- \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= - \frac{π}{6} + πn

x \geq - \frac{π}{6} + πn

x \geq - \frac{π}{6} + πn

x \geq - \frac{π}{6} + πn

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

2 x \leq arccos (\frac{1}{2}) + 2 πn

化简

arccos (\frac{1}{2}) + 2 πn : \frac{π}{3} + 2 πn

arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

2 x \leq \frac{π}{3} + 2 πn

两边除以

2

2 x \leq \frac{π}{3} + 2 πn

两边除以

2

\frac{2 x}{2} \leq \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \leq \frac{\frac{π}{3}}{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}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x \leq \frac{π}{6} + πn

x \leq \frac{π}{6} + πn

x \leq \frac{π}{6} + πn

合并区间

x \geq - \frac{π}{6} + πn and x \leq \frac{π}{6} + πn

合并重叠的区间

- \frac{π}{6} + πn \leq x \leq \frac{π}{6} + πn

合并区间

- \frac{π}{6} + πn \leq x \leq \frac{π}{6} + πn and 0 \leq x < 2 π

0 \leq x \leq \frac{π}{6} or π - \frac{π}{6} \leq x \leq π + \frac{π}{6} or 2 π - \frac{π}{6} \leq x < 2 π

作图

流行的例子

sin (2 x) > cos (x)

cos (x) \leq 1

sec^{2} (x) \leq \frac{4}{3}

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

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

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