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受欢迎的三角函数 >

2sin^2(4x)>= 0.5

解答

2 sin^{2} (4 x) \geq 0.5

解答

\frac{π}{24} + \frac{π}{2} n \leq x \leq \frac{5 π}{24} + \frac{π}{2} n or - \frac{5 π}{24} + \frac{π}{2} n \leq x \leq - \frac{π}{24} + \frac{π}{2} n

+2

间隔符号

[\frac{π}{24} + \frac{π}{2} n, \frac{5 π}{24} + \frac{π}{2} n] \cup [- \frac{5 π}{24} + \frac{π}{2} n, - \frac{π}{24} + \frac{π}{2} n]

十进制

0.13089 \dots + \frac{π}{2} n \leq x \leq 0.65449 \dots + \frac{π}{2} n or - 0.65449 \dots + \frac{π}{2} n \leq x \leq - 0.13089 \dots + \frac{π}{2} n

求解步骤

2 sin^{2} (4 x) \geq 0.5

两边除以

2

2 sin^{2} (4 x) \geq 0.5

两边除以

2

\frac{2 sin ^{2} ( 4 x )}{2} \geq \frac{0.5}{2}

化简

sin^{2} (4 x) \geq 0.25

sin^{2} (4 x) \geq 0.25

对于

u^{n} \geq a

，若

n

为偶数，则

u \leq - n a or u \geq n a

sin (4 x) \leq - 0.25 or sin (4 x) \geq 0.25

0.25 = 0.5

0.25

0.25 = 0.5

= 0.5

sin (4 x) \leq - 0.5 or sin (4 x) \geq 0.5

sin (4 x) \leq - 0.5 : - \frac{5 π}{24} + \frac{π}{2} n \leq x \leq - \frac{π}{24} + \frac{π}{2} n

sin (4 x) \leq - 0.5

对于

sin (x) \leq a

，若

- 1 < a < 1

，则

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

- π - arcsin (- 0.5) + 2 πn \leq 4 x \leq arcsin (- 0.5) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- π - arcsin (- 0.5) + 2 πn \leq 4 x and 4 x \leq arcsin (- 0.5) + 2 πn

- π - arcsin (- 0.5) + 2 πn \leq 4 x : x \geq - \frac{5 π}{24} + \frac{π}{2} n

- π - arcsin (- 0.5) + 2 πn \leq 4 x

交换两边

4 x \geq - π - arcsin (- 0.5) + 2 πn

化简

- π - arcsin (- 0.5) + 2 πn : - π + \frac{π}{6} + 2 πn

- π - arcsin (- 0.5) + 2 πn

arcsin (- 0.5) = - \frac{π}{6}

arcsin (- 0.5)

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

利用以下特性:

arcsin (- x) = - arcsin (x)

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

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

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= - π - (- \frac{π}{6}) + 2 πn

使用法则

- (- a) = a

= - π + \frac{π}{6} + 2 πn

4 x \geq - π + \frac{π}{6} + 2 πn

两边除以

4

4 x \geq - π + \frac{π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} \geq - \frac{π}{4} + \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \geq - \frac{π}{4} + \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

- \frac{π}{4} + \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : - \frac{π}{4} + \frac{π}{24} + \frac{πn}{2}

- \frac{π}{4} + \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= - \frac{π}{4} + \frac{π}{24} + \frac{πn}{2}

x \geq - \frac{π}{4} + \frac{π}{24} + \frac{πn}{2}

x \geq - \frac{π}{4} + \frac{π}{24} + \frac{πn}{2}

化简

- \frac{π}{4} + \frac{π}{24} : - \frac{5 π}{24}

- \frac{π}{4} + \frac{π}{24}

4, 24

的最小公倍数:

24

4, 24

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

24

质因数分解:

2 \cdot 2 \cdot 2 \cdot 3

24

24

除以

224 = 12 \cdot 2

= 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3

将每个因子乘以它在

4

或

24

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 2 \cdot 3 = 24

= 24

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

24

对于

\frac{π}{4} :

将分母和分子乘以

6

\frac{π}{4} = \frac{π 6}{4 \cdot 6} = \frac{π 6}{24}

= - \frac{π 6}{24} + \frac{π}{24}

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}{24}

同类项相加:

- 6 π + π = - 5 π

= \frac{- 5 π}{24}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{5 π}{24}

x \geq - \frac{5 π}{24} + \frac{π}{2} n

x \geq - \frac{5 π}{24} + \frac{π}{2} n

4 x \leq arcsin (- 0.5) + 2 πn : x \leq - \frac{π}{24} + \frac{πn}{2}

4 x \leq arcsin (- 0.5) + 2 πn

化简

arcsin (- 0.5) + 2 πn : - \frac{π}{6} + 2 πn

arcsin (- 0.5) + 2 πn

arcsin (- 0.5) = - \frac{π}{6}

arcsin (- 0.5)

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

利用以下特性:

arcsin (- x) = - arcsin (x)

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

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

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= - \frac{π}{6} + 2 πn

4 x \leq - \frac{π}{6} + 2 πn

两边除以

4

4 x \leq - \frac{π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} \leq - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \leq - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

- \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : - \frac{π}{24} + \frac{πn}{2}

- \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= - \frac{π}{24} + \frac{πn}{2}

x \leq - \frac{π}{24} + \frac{πn}{2}

x \leq - \frac{π}{24} + \frac{πn}{2}

x \leq - \frac{π}{24} + \frac{πn}{2}

合并区间

x \geq - \frac{5 π}{24} + \frac{π}{2} n and x \leq - \frac{π}{24} + \frac{πn}{2}

合并重叠的区间

- \frac{5 π}{24} + \frac{π}{2} n \leq x \leq - \frac{π}{24} + \frac{π}{2} n

sin (4 x) \geq 0.5 : \frac{π}{24} + \frac{π}{2} n \leq x \leq \frac{5 π}{24} + \frac{π}{2} n

sin (4 x) \geq 0.5

对于

sin (x) \geq a

，若

- 1 < a < 1

，则

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

arcsin (0.5) + 2 πn \leq 4 x \leq π - arcsin (0.5) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

arcsin (0.5) + 2 πn \leq 4 x and 4 x \leq π - arcsin (0.5) + 2 πn

arcsin (0.5) + 2 πn \leq 4 x : x \geq \frac{π}{24} + \frac{πn}{2}

arcsin (0.5) + 2 πn \leq 4 x

交换两边

4 x \geq arcsin (0.5) + 2 πn

化简

arcsin (0.5) + 2 πn : \frac{π}{6} + 2 πn

arcsin (0.5) + 2 πn

arcsin (0.5) = \frac{π}{6}

arcsin (0.5)

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

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= \frac{π}{6}

= \frac{π}{6} + 2 πn

4 x \geq \frac{π}{6} + 2 πn

两边除以

4

4 x \geq \frac{π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} \geq \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \geq \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π}{24} + \frac{πn}{2}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{24} + \frac{πn}{2}

x \geq \frac{π}{24} + \frac{πn}{2}

x \geq \frac{π}{24} + \frac{πn}{2}

x \geq \frac{π}{24} + \frac{πn}{2}

4 x \leq π - arcsin (0.5) + 2 πn : x \leq \frac{5 π}{24} + \frac{π}{2} n

4 x \leq π - arcsin (0.5) + 2 πn

化简

π - arcsin (0.5) + 2 πn : π - \frac{π}{6} + 2 πn

π - arcsin (0.5) + 2 πn

arcsin (0.5) = \frac{π}{6}

arcsin (0.5)

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

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= \frac{π}{6}

= π - \frac{π}{6} + 2 πn

4 x \leq π - \frac{π}{6} + 2 πn

两边除以

4

4 x \leq π - \frac{π}{6} + 2 πn

两边除以

4

\frac{4 x}{4} \leq \frac{π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} \leq \frac{π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} : \frac{π}{4} - \frac{π}{24} + \frac{πn}{2}

\frac{π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{4} - \frac{π}{24} + \frac{πn}{2}

x \leq \frac{π}{4} - \frac{π}{24} + \frac{πn}{2}

x \leq \frac{π}{4} - \frac{π}{24} + \frac{πn}{2}

化简

\frac{π}{4} - \frac{π}{24} : \frac{5 π}{24}

\frac{π}{4} - \frac{π}{24}

4, 24

的最小公倍数:

24

4, 24

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

24

质因数分解:

2 \cdot 2 \cdot 2 \cdot 3

24

24

除以

224 = 12 \cdot 2

= 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3

将每个因子乘以它在

4

或

24

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 2 \cdot 3 = 24

= 24

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

24

对于

\frac{π}{4} :

将分母和分子乘以

6

\frac{π}{4} = \frac{π 6}{4 \cdot 6} = \frac{π 6}{24}

= \frac{π 6}{24} - \frac{π}{24}

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}{24}

同类项相加:

6 π - π = 5 π

= \frac{5 π}{24}

x \leq \frac{5 π}{24} + \frac{π}{2} n

x \leq \frac{5 π}{24} + \frac{π}{2} n

合并区间

x \geq \frac{π}{24} + \frac{πn}{2} and x \leq \frac{5 π}{24} + \frac{π}{2} n

合并重叠的区间

\frac{π}{24} + \frac{π}{2} n \leq x \leq \frac{5 π}{24} + \frac{π}{2} n

合并区间

- \frac{5 π}{24} + \frac{π}{2} n \leq x \leq - \frac{π}{24} + \frac{π}{2} n or \frac{π}{24} + \frac{π}{2} n \leq x \leq \frac{5 π}{24} + \frac{π}{2} n

合并重叠的区间

\frac{π}{24} + \frac{π}{2} n \leq x \leq \frac{5 π}{24} + \frac{π}{2} n or - \frac{5 π}{24} + \frac{π}{2} n \leq x \leq - \frac{π}{24} + \frac{π}{2} n

流行的例子

cos (x) > - 1

2(cos(3x))^2+sqrt(3)sin(6x)< 1/2

2 (cos (3 x))^{2} + 3 sin (6 x) < \frac{1}{2}

sin (3 x) \leq \frac{1}{3}

tan(t)<-1/(sqrt(3))

tan (t) < - \frac{1}{3}

sin(x)>= 1/2 ,0<= x<= 2pi

sin (x) \geq \frac{1}{2}, 0 \leq x \leq 2 π