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受欢迎的三角函数 >

2cos(3x-1/2)>= sqrt(2)

解答

2 cos (3 x - \frac{1}{2}) \geq 2

解答

\frac{- π + 2}{12} + \frac{2 π}{3} n \leq x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

+2

间隔符号

[\frac{- π + 2}{12} + \frac{2 π}{3} n, \frac{π + 2}{12} + \frac{2 π}{3} n]

十进制

- 0.09513 \dots + \frac{2 π}{3} n \leq x \leq 0.42846 \dots + \frac{2 π}{3} n

求解步骤

2 cos (3 x - \frac{1}{2}) \geq 2

两边除以

2

2 cos (3 x - \frac{1}{2}) \geq 2

两边除以

2

\frac{2 cos ( 3 x - \frac{1}{2} )}{2} \geq \frac{2}{2}

化简

cos (3 x - \frac{1}{2}) \geq \frac{2}{2}

cos (3 x - \frac{1}{2}) \geq \frac{2}{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{2}{2}) + 2 πn \leq (3 x - \frac{1}{2}) \leq arccos (\frac{2}{2}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2} and 3 x - \frac{1}{2} \leq arccos (\frac{2}{2}) + 2 πn

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2} : x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

- arccos (\frac{2}{2}) + 2 πn \leq 3 x - \frac{1}{2}

交换两边

3 x - \frac{1}{2} \geq - 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

3 x - \frac{1}{2} \geq - \frac{π}{4} + 2 πn

将

\frac{1}{2}

到右边

3 x - \frac{1}{2} \geq - \frac{π}{4} + 2 πn

两边加上

\frac{1}{2}

3 x - \frac{1}{2} + \frac{1}{2} \geq - \frac{π}{4} + 2 πn + \frac{1}{2}

化简

3 x - \frac{1}{2} + \frac{1}{2} \geq - \frac{π}{4} + 2 πn + \frac{1}{2}

化简

3 x - \frac{1}{2} + \frac{1}{2} : 3 x

3 x - \frac{1}{2} + \frac{1}{2}

同类项相加:

- \frac{1}{2} + \frac{1}{2} \geq 0

= 3 x

化简

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

- \frac{π}{4} + 2 πn + \frac{1}{2}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{1}{2}

无法进一步化简

= 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

两边除以

3

3 x \geq 2 πn - \frac{π}{4} + \frac{1}{2}

两边除以

3

\frac{3 x}{3} \geq \frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

化简

\frac{3 x}{3} \geq \frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3} : \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

\frac{2 πn}{3} - \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

使用分式法则:

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

= \frac{π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{π}{12}

\frac{\frac{1}{2}}{3} = \frac{1}{6}

\frac{\frac{1}{2}}{3}

使用分式法则:

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

= \frac{1}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{1}{6}

= \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

x \geq \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

x \geq \frac{2 πn}{3} - \frac{π}{12} + \frac{1}{6}

化简

- \frac{π}{12} + \frac{1}{6} : \frac{- π + 2}{12}

- \frac{π}{12} + \frac{1}{6}

12, 6

的最小公倍数:

12

12, 6

最小公倍数 (LCM)

12

质因数分解:

2 \cdot 2 \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 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 3

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

12

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

对于

\frac{1}{6} :

将分母和分子乘以

2

\frac{1}{6} = \frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}

= - \frac{π}{12} + \frac{2}{12}

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 + 2}{12}

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n

3 x - \frac{1}{2} \leq arccos (\frac{2}{2}) + 2 πn : x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

3 x - \frac{1}{2} \leq 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

3 x - \frac{1}{2} \leq \frac{π}{4} + 2 πn

将

\frac{1}{2}

到右边

3 x - \frac{1}{2} \leq \frac{π}{4} + 2 πn

两边加上

\frac{1}{2}

3 x - \frac{1}{2} + \frac{1}{2} \leq \frac{π}{4} + 2 πn + \frac{1}{2}

化简

3 x - \frac{1}{2} + \frac{1}{2} \leq \frac{π}{4} + 2 πn + \frac{1}{2}

化简

3 x - \frac{1}{2} + \frac{1}{2} : 3 x

3 x - \frac{1}{2} + \frac{1}{2}

同类项相加:

- \frac{1}{2} + \frac{1}{2} \leq 0

= 3 x

化简

\frac{π}{4} + 2 πn + \frac{1}{2} : 2 πn + \frac{π}{4} + \frac{1}{2}

\frac{π}{4} + 2 πn + \frac{1}{2}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{1}{2}

无法进一步化简

= 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

两边除以

3

3 x \leq 2 πn + \frac{π}{4} + \frac{1}{2}

两边除以

3

\frac{3 x}{3} \leq \frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

化简

\frac{3 x}{3} \leq \frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3} : \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

\frac{2 πn}{3} + \frac{\frac{π}{4}}{3} + \frac{\frac{1}{2}}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

使用分式法则:

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

= \frac{π}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{π}{12}

\frac{\frac{1}{2}}{3} = \frac{1}{6}

\frac{\frac{1}{2}}{3}

使用分式法则:

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

= \frac{1}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{1}{6}

= \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

x \leq \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

x \leq \frac{2 πn}{3} + \frac{π}{12} + \frac{1}{6}

化简

\frac{π}{12} + \frac{1}{6} : \frac{π + 2}{12}

\frac{π}{12} + \frac{1}{6}

12, 6

的最小公倍数:

12

12, 6

最小公倍数 (LCM)

12

质因数分解:

2 \cdot 2 \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 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 3

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

12

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

对于

\frac{1}{6} :

将分母和分子乘以

2

\frac{1}{6} = \frac{1 \cdot 2}{6 \cdot 2} = \frac{2}{12}

= \frac{π}{12} + \frac{2}{12}

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 + 2}{12}

x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

合并区间

x \geq \frac{- π + 2}{12} + \frac{2 π}{3} n and x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

合并重叠的区间

\frac{- π + 2}{12} + \frac{2 π}{3} n \leq x \leq \frac{π + 2}{12} + \frac{2 π}{3} n

流行的例子

sin(θ)>0,tan(θ)<0

sin (θ) > 0, tan (θ) < 0

sin(2x)>(sqrt(2))/2

sin (2 x) > \frac{2}{2}

2 sin (x) > 1

((2cos(x)+1))/(2sin(x)-sqrt(3))>0[0.2pi]

\frac{( 2 cos ( x ) + 1 )}{2 sin ( x ) - 3} > 0 [0.2 π]

0.5sin(2t)+1.2>1.45

0.5 sin (2 t) + 1.2 > 1.45