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受欢迎的三角函数 >

2cos^2(2x)<= 0.5

解答

2 cos^{2} (2 x) \leq 0.5

解答

\frac{π}{6} + πn \leq x \leq \frac{π}{3} + πn or \frac{2 π}{3} + πn \leq x \leq \frac{5 π}{6} + πn

+2

间隔符号

[\frac{π}{6} + πn, \frac{π}{3} + πn] \cup [\frac{2 π}{3} + πn, \frac{5 π}{6} + πn]

十进制

0.52359 \dots + πn \leq x \leq 1.04719 \dots + πn or 2.09439 \dots + πn \leq x \leq 2.61799 \dots + πn

求解步骤

2 cos^{2} (2 x) \leq 0.5

两边除以

2

2 cos^{2} (2 x) \leq 0.5

两边除以

2

\frac{2 cos ^{2} ( 2 x )}{2} \leq \frac{0.5}{2}

化简

cos^{2} (2 x) \leq 0.25

cos^{2} (2 x) \leq 0.25

对于

u^{n} \leq a

，若

n

为偶数，则

- n a \leq u \leq n a

- 0.25 \leq cos (2 x) \leq 0.25

0.25 = 0.5

0.25

0.25 = 0.5

= 0.5

- 0.5 \leq cos (2 x) \leq 0.5

若

a \leq u \leq b

，则

a \leq u and u \leq b

- 0.5 \leq cos (2 x) and cos (2 x) \leq 0.5

- 0.5 \leq cos (2 x) : - \frac{π}{3} + πn \leq x \leq \frac{π}{3} + πn

- 0.5 \leq cos (2 x)

交换两边

cos (2 x) \geq - 0.5

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (- 0.5) + 2 πn \leq 2 x \leq arccos (- 0.5) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (- 0.5) + 2 πn \leq 2 x and 2 x \leq arccos (- 0.5) + 2 πn

- arccos (- 0.5) + 2 πn \leq 2 x : x \geq - \frac{π}{3} + πn

- arccos (- 0.5) + 2 πn \leq 2 x

交换两边

2 x \geq - arccos (- 0.5) + 2 πn

化简

- arccos (- 0.5) + 2 πn : - \frac{2 π}{3} + 2 πn

- arccos (- 0.5) + 2 πn

arccos (- 0.5) = \frac{2 π}{3}

arccos (- 0.5)

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

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

arccos (- \frac{1}{2})

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{2 π}{3}

= \frac{2 π}{3}

= - \frac{2 π}{3} + 2 πn

2 x \geq - \frac{2 π}{3} + 2 πn

两边除以

2

2 x \geq - \frac{2 π}{3} + 2 πn

两边除以

2

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

化简

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

使用分式法则:

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

= \frac{2 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π}{6}

约分:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= - \frac{π}{3} + πn

x \geq - \frac{π}{3} + πn

x \geq - \frac{π}{3} + πn

x \geq - \frac{π}{3} + πn

2 x \leq arccos (- 0.5) + 2 πn : x \leq \frac{π}{3} + πn

2 x \leq arccos (- 0.5) + 2 πn

化简

arccos (- 0.5) + 2 πn : \frac{2 π}{3} + 2 πn

arccos (- 0.5) + 2 πn

arccos (- 0.5) = \frac{2 π}{3}

arccos (- 0.5)

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

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

arccos (- \frac{1}{2})

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{2 π}{3}

= \frac{2 π}{3}

= \frac{2 π}{3} + 2 πn

2 x \leq \frac{2 π}{3} + 2 πn

两边除以

2

2 x \leq \frac{2 π}{3} + 2 πn

两边除以

2

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

化简

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

使用分式法则:

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

= \frac{2 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π}{6}

约分:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

x \leq \frac{π}{3} + πn

x \leq \frac{π}{3} + πn

x \leq \frac{π}{3} + πn

合并区间

x \geq - \frac{π}{3} + πn and x \leq \frac{π}{3} + πn

合并重叠的区间

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

cos (2 x) \leq 0.5 : \frac{π}{6} + πn \leq x \leq \frac{5 π}{6} + πn

cos (2 x) \leq 0.5

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

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

arccos (0.5) + 2 πn \leq 2 x \leq 2 π - arccos (0.5) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arccos (0.5) + 2 πn \leq 2 x and 2 x \leq 2 π - arccos (0.5) + 2 πn

arccos (0.5) + 2 πn \leq 2 x : x \geq \frac{π}{6} + πn

arccos (0.5) + 2 πn \leq 2 x

交换两边

2 x \geq arccos (0.5) + 2 πn

化简

arccos (0.5) + 2 πn : \frac{π}{3} + 2 πn

arccos (0.5) + 2 πn

arccos (0.5) = \frac{π}{3}

arccos (0.5)

= arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

arccos (\frac{1}{2})

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}

= \frac{π}{3}

= \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 2 π - arccos (0.5) + 2 πn : x \leq \frac{5 π}{6} + πn

2 x \leq 2 π - arccos (0.5) + 2 πn

化简

2 π - arccos (0.5) + 2 πn : 2 π - \frac{π}{3} + 2 πn

2 π - arccos (0.5) + 2 πn

arccos (0.5) = \frac{π}{3}

arccos (0.5)

= arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

arccos (\frac{1}{2})

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}

= \frac{π}{3}

= 2 π - \frac{π}{3} + 2 πn

2 x \leq 2 π - \frac{π}{3} + 2 πn

两边除以

2

2 x \leq 2 π - \frac{π}{3} + 2 πn

两边除以

2

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

化简

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

\frac{2 π}{2} - \frac{\frac{π}{3}}{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}{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

= π - \frac{π}{6} + πn

x \leq π - \frac{π}{6} + πn

x \leq π - \frac{π}{6} + πn

化简

π - \frac{π}{6} : \frac{5 π}{6}

π - \frac{π}{6}

将项转换为分式:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{π}{6}

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 6 - π}{6}

同类项相加:

6 π - π = 5 π

= \frac{5 π}{6}

x \leq \frac{5 π}{6} + πn

x \leq \frac{5 π}{6} + πn

合并区间

x \geq \frac{π}{6} + πn and x \leq \frac{5 π}{6} + πn

合并重叠的区间

\frac{π}{6} + πn \leq x \leq \frac{5 π}{6} + πn

合并区间

- \frac{π}{3} + πn \leq x \leq \frac{π}{3} + πn and \frac{π}{6} + πn \leq x \leq \frac{5 π}{6} + πn

合并重叠的区间

\frac{π}{6} + πn \leq x \leq \frac{π}{3} + πn or \frac{2 π}{3} + πn \leq x \leq \frac{5 π}{6} + πn

流行的例子

sin (2 x) < 0

cos(x)(cos(x)+2)<= 0

cos (x) (cos (x) + 2) \leq 0

(sin(x))/(cos(x))>= 1

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

cos^2(x)-sin^2(x)>= 0

cos^{2} (x) - sin^{2} (x) \geq 0

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

2 cos^{2} (x) + 3 sin (x) - 3 > 0