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受欢迎的三角函数 >

sin^2(2x)<= 1/2

解答

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

解答

πn \leq x \leq \frac{π}{8} + πn or \frac{3 π}{8} + πn \leq x \leq \frac{5 π}{8} + πn or \frac{7 π}{8} + πn \leq x < π + πn

+2

间隔符号

[πn, \frac{π}{8} + πn] \cup [\frac{3 π}{8} + πn, \frac{5 π}{8} + πn] \cup [\frac{7 π}{8} + πn, π + πn)

十进制

πn \leq x \leq 0.39269 \dots + πn or 1.17809 \dots + πn \leq x \leq 1.96349 \dots + πn or 2.74889 \dots + πn \leq x < 3.14159 \dots + πn

求解步骤

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

对于

u^{n} \leq a

，若

n

为偶数，则

- n a \leq u \leq n a

- \frac{1}{2} \leq sin (2 x) \leq \frac{1}{2}

若

a \leq u \leq b

，则

a \leq u and u \leq b

- \frac{1}{2} \leq sin (2 x) and sin (2 x) \leq \frac{1}{2}

- \frac{1}{2} \leq sin (2 x) : - \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn

- \frac{1}{2} \leq sin (2 x)

交换两边

sin (2 x) \geq - \frac{1}{2}

对于

sin (x) \geq a

，若

- 1 < a < 1

，则

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

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

arcsin (- \frac{1}{2}) + 2 πn \leq 2 x : x \geq - \frac{π}{8} + πn

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

交换两边

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

化简

arcsin (- \frac{1}{2}) + 2 πn : - \frac{π}{4} + 2 πn

arcsin (- \frac{1}{2}) + 2 πn

arcsin (- \frac{1}{2}) = - \frac{π}{4}

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}{4}

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}{4}

= - \frac{π}{4}

= - \frac{π}{4} + 2 πn

2 x \geq - \frac{π}{4} + 2 πn

两边除以

2

2 x \geq - \frac{π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} \geq - \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 \geq - \frac{π}{8} + πn

x \geq - \frac{π}{8} + πn

x \geq - \frac{π}{8} + πn

2 x \leq π - arcsin (- \frac{1}{2}) + 2 πn : x \leq \frac{5 π}{8} + πn

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

化简

π - arcsin (- \frac{1}{2}) + 2 πn : π + \frac{π}{4} + 2 πn

π - arcsin (- \frac{1}{2}) + 2 πn

arcsin (- \frac{1}{2}) = - \frac{π}{4}

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}{4}

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}{4}

= - \frac{π}{4}

= π - (- \frac{π}{4}) + 2 πn

使用法则

- (- a) = a

= π + \frac{π}{4} + 2 πn

2 x \leq π + \frac{π}{4} + 2 πn

两边除以

2

2 x \leq π + \frac{π}{4} + 2 πn

两边除以

2

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

化简

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

\frac{π}{2} + \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}{2} + \frac{π}{8} + πn

x \leq \frac{π}{2} + \frac{π}{8} + πn

x \leq \frac{π}{2} + \frac{π}{8} + πn

化简

\frac{π}{2} + \frac{π}{8} : \frac{5 π}{8}

\frac{π}{2} + \frac{π}{8}

2, 8

的最小公倍数:

8

2, 8

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

8

质因数分解:

2 \cdot 2 \cdot 2

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

将每个因子乘以它在

2

或

8

中出现的最多次数

= 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 = 8

= 8

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

8

对于

\frac{π}{2} :

将分母和分子乘以

4

\frac{π}{2} = \frac{π 4}{2 \cdot 4} = \frac{π 4}{8}

= \frac{π 4}{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 4 + π}{8}

同类项相加:

4 π + π = 5 π

= \frac{5 π}{8}

x \leq \frac{5 π}{8} + πn

x \leq \frac{5 π}{8} + πn

合并区间

x \geq - \frac{π}{8} + πn and x \leq \frac{5 π}{8} + πn

合并重叠的区间

- \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn

sin (2 x) \leq \frac{1}{2} : - \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

sin (2 x) \leq \frac{1}{2}

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

- π - arcsin (\frac{1}{2}) + 2 πn \leq 2 x : x \geq - \frac{5 π}{8} + πn

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

交换两边

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

化简

- π - arcsin (\frac{1}{2}) + 2 πn : - π - \frac{π}{4} + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{4}

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}

= - π - \frac{π}{4} + 2 πn

2 x \geq - π - \frac{π}{4} + 2 πn

两边除以

2

2 x \geq - π - \frac{π}{4} + 2 πn

两边除以

2

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

化简

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

- \frac{π}{2} - \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}{2} - \frac{π}{8} + πn

x \geq - \frac{π}{2} - \frac{π}{8} + πn

x \geq - \frac{π}{2} - \frac{π}{8} + πn

化简

- \frac{π}{2} - \frac{π}{8} : - \frac{5 π}{8}

- \frac{π}{2} - \frac{π}{8}

2, 8

的最小公倍数:

8

2, 8

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

8

质因数分解:

2 \cdot 2 \cdot 2

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

将每个因子乘以它在

2

或

8

中出现的最多次数

= 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 = 8

= 8

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

8

对于

\frac{π}{2} :

将分母和分子乘以

4

\frac{π}{2} = \frac{π 4}{2 \cdot 4} = \frac{π 4}{8}

= - \frac{π 4}{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 4 - π}{8}

同类项相加:

- 4 π - π = - 5 π

= \frac{- 5 π}{8}

使用分式法则:

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

= - \frac{5 π}{8}

x \geq - \frac{5 π}{8} + πn

x \geq - \frac{5 π}{8} + πn

2 x \leq arcsin (\frac{1}{2}) + 2 πn : x \leq \frac{π}{8} + πn

2 x \leq arcsin (\frac{1}{2}) + 2 πn

化简

arcsin (\frac{1}{2}) + 2 πn : \frac{π}{4} + 2 πn

arcsin (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{4}

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}{4} + 2 πn

2 x \leq \frac{π}{4} + 2 πn

两边除以

2

2 x \leq \frac{π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} \leq \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 \leq \frac{π}{8} + πn

x \leq \frac{π}{8} + πn

x \leq \frac{π}{8} + πn

合并区间

x \geq - \frac{5 π}{8} + πn and x \leq \frac{π}{8} + πn

合并重叠的区间

- \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

合并区间

- \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn and - \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

合并重叠的区间

πn \leq x \leq \frac{π}{8} + πn or \frac{3 π}{8} + πn \leq x \leq \frac{5 π}{8} + πn or \frac{7 π}{8} + πn \leq x < π + πn

流行的例子

(1-2cos^2(x))/(tan(x))>0

\frac{1 - 2 cos ^{2} ( x )}{tan ( x )} > 0

2cos^2(x)+cos(x)>0

2 cos^{2} (x) + cos (x) > 0

tan(2x)<= sqrt(3)

tan (2 x) \leq 3

(2sin(θ)cos(θ))/((3cos^2(θ)+1))>= 16/45

\frac{2 sin ( θ ) cos ( θ )}{( 3 cos ^{2} ( θ ) + 1 )} \geq \frac{16}{45}

cos (2 t) \geq - \frac{1}{2}