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受欢迎的三角函数 >

5-8cos(x)+4cos(2x)<0

解答

5 - 8 cos (x) + 4 cos (2 x) < 0

解答

arccos (\frac{2 + 2}{4}) + 2 πn < x < arccos (\frac{- 2 + 2}{4}) + 2 πn or - arccos (\frac{- 2 + 2}{4}) + 2 π + 2 πn < x < 2 π - arccos (\frac{2 + 2}{4}) + 2 πn

+2

间隔符号

(arccos (\frac{2 + 2}{4}) + 2 πn, arccos (\frac{- 2 + 2}{4}) + 2 πn) \cup (- arccos (\frac{- 2 + 2}{4}) + 2 π + 2 πn, 2 π - arccos (\frac{2 + 2}{4}) + 2 πn)

十进制

0.54802 \dots + 2 πn < x < 1.42382 \dots + 2 πn or 4.85936 \dots + 2 πn < x < 5.73515 \dots + 2 πn

求解步骤

5 - 8 cos (x) + 4 cos (2 x) < 0

利用以下特性:

cos (2 x) = - 1 + 2 cos^{2} (x)

5 + 4 (- 1 + 2 cos^{2} (x)) - 8 cos (x) < 0

化简

5 + 4 (- 1 + 2 cos^{2} (x)) - 8 cos (x) : 8 cos^{2} (x) - 8 cos (x) + 1

5 + 4 (- 1 + 2 cos^{2} (x)) - 8 cos (x)

乘开

4 (- 1 + 2 cos^{2} (x)) : - 4 + 8 cos^{2} (x)

4 (- 1 + 2 cos^{2} (x))

使用分配律:

a (b + c) = ab + a c

a = 4, b = - 1, c = 2 cos^{2} (x)

= 4 (- 1) + 4 \cdot 2 cos^{2} (x)

使用加减运算法则

+ (- a) = - a

= - 4 \cdot 1 + 4 \cdot 2 cos^{2} (x)

化简

- 4 \cdot 1 + 4 \cdot 2 cos^{2} (x) : - 4 + 8 cos^{2} (x)

- 4 \cdot 1 + 4 \cdot 2 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 cos^{2} (x)

数字相乘:

4 \cdot 2 = 8

= - 4 + 8 cos^{2} (x)

= - 4 + 8 cos^{2} (x)

= 5 - 4 + 8 cos^{2} (x) - 8 cos (x)

数字相减:

5 - 4 = 1

= 8 cos^{2} (x) - 8 cos (x) + 1

8 cos^{2} (x) - 8 cos (x) + 1 < 0

令

: u = cos (x)

8 u^{2} - 8 u + 1 < 0

8 u^{2} - 8 u + 1 < 0 : - \frac{2}{4} + \frac{1}{2} < u < \frac{2}{4} + \frac{1}{2}

8 u^{2} - 8 u + 1 < 0

对

8 u^{2} - 8 u + 1

配方:

8 (u - \frac{1}{2})^{2} - 1

8 u^{2} - 8 u + 1

将

8 u^{2} - 8 u + 1

改写为如下形式:

x^{2} + 2 a x + a^{2}

因式分解

8

8 (u^{2} - u + \frac{1}{8})

2 a = - 1 : a = - \frac{1}{2}

2 a = - 1

两边除以

2

2 a = - 1

两边除以

2

\frac{2 a}{2} = \frac{- 1}{2}

化简

a = - \frac{1}{2}

a = - \frac{1}{2}

加和减

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

8 (u^{2} - u + \frac{1}{8} + (- \frac{1}{2})^{2} - (- \frac{1}{2})^{2})

x^{2} + 2 a x + a^{2} = (x + a)^{2}

u^{2} - 1 u + (- \frac{1}{2})^{2} = (u - \frac{1}{2})^{2}

8 ((u - \frac{1}{2})^{2} + \frac{1}{8} - (- \frac{1}{2})^{2})

化简

8 (u - \frac{1}{2})^{2} - 1

8 (u - \frac{1}{2})^{2} - 1 < 0

将

1

到右边

8 (u - \frac{1}{2})^{2} - 1 < 0

两边加上

1

8 (u - \frac{1}{2})^{2} - 1 + 1 < 0 + 1

化简

8 (u - \frac{1}{2})^{2} < 1

8 (u - \frac{1}{2})^{2} < 1

两边除以

8

8 (u - \frac{1}{2})^{2} < 1

两边除以

8

\frac{8 ( u - \frac{1}{2} ) ^{2}}{8} < \frac{1}{8}

化简

(u - \frac{1}{2})^{2} < \frac{1}{8}

(u - \frac{1}{2})^{2} < \frac{1}{8}

对于

u^{n} < a

，若

n

为偶数，则

- n a < u < n a

- \frac{1}{8} < u - \frac{1}{2} < \frac{1}{8}

若

a < u < b

，则

a < u and u < b

- \frac{1}{8} < u - \frac{1}{2} and u - \frac{1}{2} < \frac{1}{8}

- \frac{1}{8} < u - \frac{1}{2} : u > - \frac{2}{4} + \frac{1}{2}

- \frac{1}{8} < u - \frac{1}{2}

交换两边

u - \frac{1}{2} > - \frac{1}{8}

化简

- \frac{1}{8} : - \frac{1}{2 2}

- \frac{1}{8}

化简

\frac{1}{8} : \frac{1}{2 2}

\frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{1}{2 2}

= - \frac{1}{2 2}

使用法则

1 = 1

= - \frac{1}{2 2}

u - \frac{1}{2} > - \frac{1}{2 2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} > - \frac{1}{2 2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} > - \frac{1}{2 2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} > - \frac{1}{2 2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} : u

u - \frac{1}{2} + \frac{1}{2}

同类项相加:

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

= u

化简

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

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

\frac{1}{2 2} = \frac{2}{4}

\frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

= 2^{1 + 2 \cdot \frac{1}{2}}

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

= - \frac{2}{4} + \frac{1}{2}

u > - \frac{2}{4} + \frac{1}{2}

u > - \frac{2}{4} + \frac{1}{2}

u > - \frac{2}{4} + \frac{1}{2}

u - \frac{1}{2} < \frac{1}{8} : u < \frac{2}{4} + \frac{1}{2}

u - \frac{1}{2} < \frac{1}{8}

化简

\frac{1}{8} : \frac{1}{2 2}

\frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{1}{2 2}

使用法则

1 = 1

= \frac{1}{2 2}

u - \frac{1}{2} < \frac{1}{2 2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} < \frac{1}{2 2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} < \frac{1}{2 2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} < \frac{1}{2 2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} : u

u - \frac{1}{2} + \frac{1}{2}

同类项相加:

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

= u

化简

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

\frac{1}{2 2} + \frac{1}{2}

\frac{1}{2 2} = \frac{2}{4}

\frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

= 2^{1 + 2 \cdot \frac{1}{2}}

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

= \frac{2}{4} + \frac{1}{2}

u < \frac{2}{4} + \frac{1}{2}

u < \frac{2}{4} + \frac{1}{2}

u < \frac{2}{4} + \frac{1}{2}

合并区间

u > - \frac{2}{4} + \frac{1}{2} and u < \frac{2}{4} + \frac{1}{2}

合并重叠的区间

u > - \frac{2}{4} + \frac{1}{2} and u < \frac{2}{4} + \frac{1}{2}

两个区间的交集是指同时存在于这两个区间的数的集合

u > - \frac{2}{4} + \frac{1}{2}

and

u < \frac{2}{4} + \frac{1}{2}

- \frac{2}{4} + \frac{1}{2} < u < \frac{2}{4} + \frac{1}{2}

- \frac{2}{4} + \frac{1}{2} < u < \frac{2}{4} + \frac{1}{2}

- \frac{2}{4} + \frac{1}{2} < u < \frac{2}{4} + \frac{1}{2}

u = cos (x)

代回

- \frac{2}{4} + \frac{1}{2} < cos (x) < \frac{2}{4} + \frac{1}{2}

若

a < u < b

，则

a < u and u < b

- \frac{2}{4} + \frac{1}{2} < cos (x) and cos (x) < \frac{2}{4} + \frac{1}{2}

- \frac{2}{4} + \frac{1}{2} < cos (x) : - arccos (\frac{- 2 + 2}{4}) + 2 πn < x < arccos (\frac{- 2 + 2}{4}) + 2 πn

- \frac{2}{4} + \frac{1}{2} < cos (x)

交换两边

cos (x) > - \frac{2}{4} + \frac{1}{2}

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- \frac{2}{4} + \frac{1}{2}) + 2 πn < x < arccos (- \frac{2}{4} + \frac{1}{2}) + 2 πn

化简

- arccos (- \frac{2}{4} + \frac{1}{2}) : - arccos (\frac{- 2 + 2}{4})

- arccos (- \frac{2}{4} + \frac{1}{2})

化简

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

- \frac{2}{4} + \frac{1}{2}

4, 2

的最小公倍数:

4

4, 2

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

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

4

或

2

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

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

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= - \frac{2}{4} + \frac{2}{4}

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{- 2 + 2}{4}

分解

- 2 + 2 : 2 (- 1 + 2)

- 2 + 2

2 = 22

= - 2 + 22

因式分解出通项

2

= 2 (- 1 + 2)

= \frac{2 ( - 1 + 2 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

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

消掉

\frac{2 ( - 1 + 2 )}{2 ^{2}} : \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

\frac{2 ( - 1 + 2 )}{2 ^{2}}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{1}{2}}}

= \frac{- 1 + 2}{2 ^{2 - \frac{1}{2}}}

数字相减:

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

= \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

= \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{- 1 + 2}{2 2}

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

= - arccos (\frac{2 - 2}{4})

化简

arccos (- \frac{2}{4} + \frac{1}{2}) : arccos (\frac{- 2 + 2}{4})

arccos (- \frac{2}{4} + \frac{1}{2})

化简

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

- \frac{2}{4} + \frac{1}{2}

4, 2

的最小公倍数:

4

4, 2

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

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

4

或

2

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

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

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= - \frac{2}{4} + \frac{2}{4}

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{- 2 + 2}{4}

分解

- 2 + 2 : 2 (- 1 + 2)

- 2 + 2

2 = 22

= - 2 + 22

因式分解出通项

2

= 2 (- 1 + 2)

= \frac{2 ( - 1 + 2 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

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

消掉

\frac{2 ( - 1 + 2 )}{2 ^{2}} : \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

\frac{2 ( - 1 + 2 )}{2 ^{2}}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{1}{2}}}

= \frac{- 1 + 2}{2 ^{2 - \frac{1}{2}}}

数字相减:

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

= \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

= \frac{- 1 + 2}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{- 1 + 2}{2 2}

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

= arccos (\frac{- 2 + 2}{4})

- arccos (\frac{- 2 + 2}{4}) + 2 πn < x < arccos (\frac{- 2 + 2}{4}) + 2 πn

cos (x) < \frac{2}{4} + \frac{1}{2} : arccos (\frac{2 + 2}{4}) + 2 πn < x < 2 π - arccos (\frac{2 + 2}{4}) + 2 πn

cos (x) < \frac{2}{4} + \frac{1}{2}

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (\frac{2}{4} + \frac{1}{2}) + 2 πn < x < 2 π - arccos (\frac{2}{4} + \frac{1}{2}) + 2 πn

化简

arccos (\frac{2}{4} + \frac{1}{2}) : arccos (\frac{2 + 2}{4})

arccos (\frac{2}{4} + \frac{1}{2})

化简

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

\frac{2}{4} + \frac{1}{2}

4, 2

的最小公倍数:

4

4, 2

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

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

4

或

2

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

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

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{2}{4}

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{2 + 2}{4}

分解

2 + 2 : 2 (1 + 2)

2 + 2

2 = 22

= 2 + 22

因式分解出通项

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 ( 1 + 2 )}{2 ^{2}}

消掉

\frac{2 ( 1 + 2 )}{2 ^{2}} : \frac{1 + 2}{2 ^{\frac{3}{2}}}

\frac{2 ( 1 + 2 )}{2 ^{2}}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( 1 + 2 )}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{1}{2}}}

= \frac{1 + 2}{2 ^{2 - \frac{1}{2}}}

数字相减:

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

= \frac{1 + 2}{2 ^{\frac{3}{2}}}

= \frac{1 + 2}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{1 + 2}{2 2}

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

= arccos (\frac{2 + 2}{4})

化简

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

2 π - arccos (\frac{2}{4} + \frac{1}{2})

化简

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

\frac{2}{4} + \frac{1}{2}

4, 2

的最小公倍数:

4

4, 2

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

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

4

或

2

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

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

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{2}{4}

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{2 + 2}{4}

= 2 π - arccos (\frac{2 + 2}{4})

arccos (\frac{2 + 2}{4}) + 2 πn < x < 2 π - arccos (\frac{2 + 2}{4}) + 2 πn

合并区间

- arccos (\frac{- 2 + 2}{4}) + 2 πn < x < arccos (\frac{- 2 + 2}{4}) + 2 πn and arccos (\frac{2 + 2}{4}) + 2 πn < x < 2 π - arccos (\frac{2 + 2}{4}) + 2 πn

合并重叠的区间

arccos (\frac{2 + 2}{4}) + 2 πn < x < arccos (\frac{- 2 + 2}{4}) + 2 πn or - arccos (\frac{- 2 + 2}{4}) + 2 π + 2 πn < x < 2 π - arccos (\frac{2 + 2}{4}) + 2 πn

流行的例子

sin (a) < 0

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

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

cos^2(x)-1/4 <= 0

cos^{2} (x) - \frac{1}{4} \leq 0

-sin(θ)+cos(θ)-sqrt(10)<0

- sin (θ) + cos (θ) - 10 < 0

cos(x)+sqrt(3)sin(x)>= sqrt(2)

cos (x) + 3 sin (x) \geq 2