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受欢迎的三角函数 >

1-4(cos(x)sin(x/2))>= 0

解答

1 - 4 (cos (x) sin (\frac{x}{2})) \geq 0

解答

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn or \frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn or 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn or x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

+2

间隔符号

(- \infty + 4 πn, 2 arcsin (\frac{5 - 1}{4}) + 4 πn] \cup [\frac{π}{3} + 4 πn, \frac{5 π}{3} + 4 πn] \cup [2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn, 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn] \cup [- 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn, \infty + 4 πn)

十进制

x \leq 0.62831 \dots + 4 πn or 1.04719 \dots + 4 πn \leq x \leq 5.23598 \dots + 4 πn or 5.65486 \dots + 4 πn \leq x \leq 8.16814 \dots + 4 πn or x \geq 10.68141 \dots + 4 πn

求解步骤

1 - 4 cos (x) sin (\frac{x}{2}) \geq 0

令

: u = \frac{x}{2}

1 - 4 cos (2 u) sin (u) \geq 0

1 - 4 cos (2 u) sin (u) \geq 0 : 2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn

1 - 4 cos (2 u) sin (u) \geq 0

利用以下特性:

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

1 - 4 (1 - 2 sin^{2} (u)) sin (u) \geq 0

令

: v = sin (u)

1 - 4 (1 - 2 v^{2}) v \geq 0

1 - 4 (1 - 2 v^{2}) v \geq 0 : - \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v \geq \frac{1}{2}

1 - 4 (1 - 2 v^{2}) v \geq 0

因式分解

1 - 4 (1 - 2 v^{2}) v : (2 v - 1) (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4})

1 - 4 (1 - 2 v^{2}) v

4 (1 - 2 v^{2}) v = - 4 v (2 v + 1) (2 v - 1)

4 (1 - 2 v^{2}) v

分解

- 2 v^{2} + 1 : - (2 v + 1) (2 v - 1)

- 2 v^{2} + 1

因式分解出通项

- 1

= - (2 v^{2} - 1)

分解

2 v^{2} - 1 : (2 v + 1) (2 v - 1)

2 v^{2} - 1

将

2 v^{2} - 1

改写为

(2 v)^{2} - 1^{2}

2 v^{2} - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} v^{2} - 1

将

1

改写为

1^{2}

= (2)^{2} v^{2} - 1^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} v^{2} = (2 v)^{2}

= (2 v)^{2} - 1^{2}

= (2 v)^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 v)^{2} - 1^{2} = (2 v + 1) (2 v - 1)

= (2 v + 1) (2 v - 1)

= - (2 v + 1) (2 v - 1)

= - 4 v (2 v + 1) (2 v - 1)

= 1 - (- 4 v (2 v + 1) (2 v - 1))

使用法则

- (- a) = a

= 1 + 4 v (2 v + 1) (2 v - 1)

乘开

1 + 4 v (2 v + 1) (2 v - 1) : 1 + 8 v^{3} - 4 v

1 + 4 v (2 v + 1) (2 v - 1)

乘开

4 v (2 v + 1) (2 v - 1) : 8 v^{3} - 4 v

乘开

(2 v + 1) (2 v - 1) : 2 v^{2} - 1

(2 v + 1) (2 v - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 v, b = 1

= (2 v)^{2} - 1^{2}

化简

(2 v)^{2} - 1^{2} : 2 v^{2} - 1

(2 v)^{2} - 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (2 v)^{2} - 1

(2 v)^{2} = 2 v^{2}

(2 v)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= (2)^{2} v^{2}

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 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

= 2 v^{2}

= 2 v^{2} - 1

= 2 v^{2} - 1

= 4 v (2 v^{2} - 1)

乘开

4 v (2 v^{2} - 1) : 8 v^{3} - 4 v

4 v (2 v^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = 4 v, b = 2 v^{2}, c = 1

= 4 v \cdot 2 v^{2} - 4 v \cdot 1

= 4 \cdot 2 v^{2} v - 4 \cdot 1 \cdot v

化简

4 \cdot 2 v^{2} v - 4 \cdot 1 \cdot v : 8 v^{3} - 4 v

4 \cdot 2 v^{2} v - 4 \cdot 1 \cdot v

4 \cdot 2 v^{2} v = 8 v^{3}

4 \cdot 2 v^{2} v

数字相乘:

4 \cdot 2 = 8

= 8 v^{2} v

使用指数法则:

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

v^{2} v = v^{2 + 1}

= 8 v^{2 + 1}

数字相加:

2 + 1 = 3

= 8 v^{3}

4 \cdot 1 \cdot v = 4 v

4 \cdot 1 \cdot v

数字相乘:

4 \cdot 1 = 4

= 4 v

= 8 v^{3} - 4 v

= 8 v^{3} - 4 v

= 8 v^{3} - 4 v

= 1 + 8 v^{3} - 4 v

= 1 + 8 v^{3} - 4 v

分解

8 v^{3} - 4 v + 1 : (2 v - 1) (4 v^{2} + 2 v - 1)

8 v^{3} - 4 v + 1

使用有理根定理

a_{0} = 1, a_{n} = 8

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4, 8

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4 , 8}

\frac{1}{2}

是表达式的根，所以因式分解

2 v - 1

= (2 v - 1) \frac{8 v ^{3} - 4 v + 1}{2 v - 1}

\frac{8 v ^{3} - 4 v + 1}{2 v - 1} = 4 v^{2} + 2 v - 1

\frac{8 v ^{3} - 4 v + 1}{2 v - 1}

对

\frac{8 v ^{3} - 4 v + 1}{2 v - 1}

做除法:

\frac{8 v ^{3} - 4 v + 1}{2 v - 1} = 4 v^{2} + \frac{4 v ^{2} - 4 v + 1}{2 v - 1}

将分子

8 v^{3} - 4 v + 1

与除数

2 v - 1

的首项系数相除：

\frac{8 v ^{3}}{2 v} = 4 v^{2}

商 = 4 v^{2}

将

2 v - 1

乘以

4 v^{2} : 8 v^{3} - 4 v^{2}

将

8 v^{3} - 4 v + 1

减去

8 v^{3} - 4 v^{2}

得到新的余数

余数 = 4 v^{2} - 4 v + 1

因此

\frac{8 v ^{3} - 4 v + 1}{2 v - 1} = 4 v^{2} + \frac{4 v ^{2} - 4 v + 1}{2 v - 1}

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

对

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

做除法:

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

将分子

4 v^{2} - 4 v + 1

与除数

2 v - 1

的首项系数相除：

\frac{4 v ^{2}}{2 v} = 2 v

商 = 2 v

将

2 v - 1

乘以

2 v : 4 v^{2} - 2 v

将

4 v^{2} - 4 v + 1

减去

4 v^{2} - 2 v

得到新的余数

余数 = - 2 v + 1

因此

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

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

对

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

做除法:

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

将分子

- 2 v + 1

与除数

2 v - 1

的首项系数相除：

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

商 = - 1

将

2 v - 1

乘以

- 1 : - 2 v + 1

将

- 2 v + 1

减去

- 2 v + 1

得到新的余数

余数 = 0

因此

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

= 4 v^{2} + 2 v - 1

= 4 v^{2} + 2 v - 1

= (2 v - 1) (4 v^{2} + 2 v - 1)

= (2 v - 1) (4 v^{2} + 2 v - 1)

因式分解

4 v^{2} + 2 v - 1 : (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4})

4 v^{2} + 2 v - 1

a x^{2} + b x + c

形式的二次多项式，若根为

x_{1}, x_{2},

，可改写为

(x - x_{1}) (x - x_{2})

4 v^{2} + 2 v - 1 = 0 : v = \frac{- 1 + 5}{4}, v = - \frac{1 + 5}{4}

4 v^{2} + 2 v - 1 = 0

使用求根公式求解

4 v^{2} + 2 v - 1 = 0

二次方程求根公式:

若

a = 4, b = 2, c = - 1

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

2^{2} - 4 \cdot 4 (- 1) = 25

2^{2} - 4 \cdot 4 (- 1)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

v_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 4}

将解分隔开

v_{1} = \frac{- 2 + 2 5}{2 \cdot 4}, v_{2} = \frac{- 2 - 2 5}{2 \cdot 4}

v = \frac{- 2 + 2 5}{2 \cdot 4} : \frac{- 1 + 5}{4}

\frac{- 2 + 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 + 2 5}{8}

分解

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

改写为

= - 2 \cdot 1 + 25

因式分解出通项

2

= 2 (- 1 + 5)

= \frac{2 ( - 1 + 5 )}{8}

约分:

2

= \frac{- 1 + 5}{4}

v = \frac{- 2 - 2 5}{2 \cdot 4} : - \frac{1 + 5}{4}

\frac{- 2 - 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{8}

分解

- 2 - 25 : - 2 (1 + 5)

- 2 - 25

改写为

= - 2 \cdot 1 - 25

因式分解出通项

2

= - 2 (1 + 5)

= - \frac{2 ( 1 + 5 )}{8}

约分:

2

= - \frac{1 + 5}{4}

二次方程组的解是:

v = \frac{- 1 + 5}{4}, v = - \frac{1 + 5}{4}

4 v^{2} + 2 v - 1 = (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4})

= (2 v - 1) (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4})

(2 v - 1) (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4}) \geq 0

确定区间

确定

(2 v - 1) (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4})

符号

确定

2 v - 1

符号

2 v - 1 = 0 : v = \frac{1}{2}

2 v - 1 = 0

将

1

到右边

2 v - 1 = 0

两边加上

1

2 v - 1 + 1 = 0 + 1

化简

2 v = 1

2 v = 1

两边除以

2

2 v = 1

两边除以

2

\frac{2 v}{2} = \frac{1}{2}

化简

v = \frac{1}{2}

v = \frac{1}{2}

2 v - 1 < 0 : v < \frac{1}{2}

2 v - 1 < 0

将

1

到右边

2 v - 1 < 0

两边加上

1

2 v - 1 + 1 < 0 + 1

化简

2 v < 1

2 v < 1

两边除以

2

2 v < 1

两边除以

2

\frac{2 v}{2} < \frac{1}{2}

化简

v < \frac{1}{2}

v < \frac{1}{2}

2 v - 1 > 0 : v > \frac{1}{2}

2 v - 1 > 0

将

1

到右边

2 v - 1 > 0

两边加上

1

2 v - 1 + 1 > 0 + 1

化简

2 v > 1

2 v > 1

两边除以

2

2 v > 1

两边除以

2

\frac{2 v}{2} > \frac{1}{2}

化简

v > \frac{1}{2}

v > \frac{1}{2}

确定

v - \frac{- 1 + 5}{4}

符号

v - \frac{- 1 + 5}{4} = 0 : v = \frac{5 - 1}{4}

v - \frac{- 1 + 5}{4} = 0

将

\frac{- 1 + 5}{4}

到右边

v - \frac{- 1 + 5}{4} = 0

两边加上

\frac{- 1 + 5}{4}

v - \frac{- 1 + 5}{4} + \frac{- 1 + 5}{4} = 0 + \frac{- 1 + 5}{4}

化简

v = \frac{5 - 1}{4}

v = \frac{5 - 1}{4}

v - \frac{- 1 + 5}{4} < 0 : v < \frac{5 - 1}{4}

v - \frac{- 1 + 5}{4} < 0

将

\frac{- 1 + 5}{4}

到右边

v - \frac{- 1 + 5}{4} < 0

两边加上

\frac{- 1 + 5}{4}

v - \frac{- 1 + 5}{4} + \frac{- 1 + 5}{4} < 0 + \frac{- 1 + 5}{4}

化简

v < \frac{5 - 1}{4}

v < \frac{5 - 1}{4}

v - \frac{- 1 + 5}{4} > 0 : v > \frac{5 - 1}{4}

v - \frac{- 1 + 5}{4} > 0

将

\frac{- 1 + 5}{4}

到右边

v - \frac{- 1 + 5}{4} > 0

两边加上

\frac{- 1 + 5}{4}

v - \frac{- 1 + 5}{4} + \frac{- 1 + 5}{4} > 0 + \frac{- 1 + 5}{4}

化简

v > \frac{5 - 1}{4}

v > \frac{5 - 1}{4}

确定

v + \frac{1 + 5}{4}

符号

v + \frac{1 + 5}{4} = 0 : v = - \frac{1 + 5}{4}

v + \frac{1 + 5}{4} = 0

将

\frac{1 + 5}{4}

到右边

v + \frac{1 + 5}{4} = 0

两边减去

\frac{1 + 5}{4}

v + \frac{1 + 5}{4} - \frac{1 + 5}{4} = 0 - \frac{1 + 5}{4}

化简

v = - \frac{1 + 5}{4}

v = - \frac{1 + 5}{4}

v + \frac{1 + 5}{4} < 0 : v < - \frac{1 + 5}{4}

v + \frac{1 + 5}{4} < 0

将

\frac{1 + 5}{4}

到右边

v + \frac{1 + 5}{4} < 0

两边减去

\frac{1 + 5}{4}

v + \frac{1 + 5}{4} - \frac{1 + 5}{4} < 0 - \frac{1 + 5}{4}

化简

v < - \frac{1 + 5}{4}

v < - \frac{1 + 5}{4}

v + \frac{1 + 5}{4} > 0 : v > - \frac{1 + 5}{4}

v + \frac{1 + 5}{4} > 0

将

\frac{1 + 5}{4}

到右边

v + \frac{1 + 5}{4} > 0

两边减去

\frac{1 + 5}{4}

v + \frac{1 + 5}{4} - \frac{1 + 5}{4} > 0 - \frac{1 + 5}{4}

化简

v > - \frac{1 + 5}{4}

v > - \frac{1 + 5}{4}

总结如下表：

2 v - 1 v - \frac{- 1 + 5}{4} v + \frac{1 + 5}{4} (2 v - 1) (v - \frac{- 1 + 5}{4}) (v + \frac{1 + 5}{4}) v < - \frac{1 + 5}{4} - - - - v = - \frac{1 + 5}{4} - - 00 - \frac{1 + 5}{4} < v < \frac{5 - 1}{4} - - + + v = \frac{5 - 1}{4} - 0 + 0 \frac{5 - 1}{4} < v < \frac{1}{2} - + + - v = \frac{1}{2} 0 + + 0 v > \frac{1}{2} + + + +

确定满足所需条件的区间：

\geq 0

v = - \frac{1 + 5}{4} or - \frac{1 + 5}{4} < v < \frac{5 - 1}{4} or v = \frac{5 - 1}{4} or v = \frac{1}{2} or v > \frac{1}{2}

合并重叠的区间

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v = \frac{1}{2} or v > \frac{1}{2}

两个区间的并集是指存在于任一区间的数的集合

v = - \frac{1 + 5}{4}

or

- \frac{1 + 5}{4} < v < \frac{5 - 1}{4}

- \frac{1 + 5}{4} \leq v < \frac{5 - 1}{4}

两个区间的并集是指存在于任一区间的数的集合

- \frac{1 + 5}{4} \leq v < \frac{5 - 1}{4}

or

v = \frac{5 - 1}{4}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4}

两个区间的并集是指存在于任一区间的数的集合

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4}

or

v = \frac{1}{2}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v = \frac{1}{2}

两个区间的并集是指存在于任一区间的数的集合

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v = \frac{1}{2}

or

v > \frac{1}{2}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v \geq \frac{1}{2}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v \geq \frac{1}{2}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v \geq \frac{1}{2}

- \frac{1 + 5}{4} \leq v \leq \frac{5 - 1}{4} or v \geq \frac{1}{2}

v = sin (u)

代回

- \frac{1 + 5}{4} \leq sin (u) \leq \frac{5 - 1}{4} or sin (u) \geq \frac{1}{2}

- \frac{1 + 5}{4} \leq sin (u) \leq \frac{5 - 1}{4} : 2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn

- \frac{1 + 5}{4} \leq sin (u) \leq \frac{5 - 1}{4}

若

a \leq u \leq b

，则

a \leq u and u \leq b

- \frac{1 + 5}{4} \leq sin (u) and sin (u) \leq \frac{5 - 1}{4}

- \frac{1 + 5}{4} \leq sin (u) : - arcsin (\frac{1 + 5}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

- \frac{1 + 5}{4} \leq sin (u)

交换两边

sin (u) \geq - \frac{1 + 5}{4}

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 + 5}{4}) + 2 πn \leq u \leq π - arcsin (- \frac{1 + 5}{4}) + 2 πn

化简

arcsin (- \frac{1 + 5}{4}) : - arcsin (\frac{1 + 5}{4})

arcsin (- \frac{1 + 5}{4})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1 + 5}{4}) = - arcsin (\frac{1 + 5}{4})

= - arcsin (\frac{1 + 5}{4})

化简

π - arcsin (- \frac{1 + 5}{4}) : π + arcsin (\frac{1 + 5}{4})

π - arcsin (- \frac{1 + 5}{4})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1 + 5}{4}) = - arcsin (\frac{1 + 5}{4})

= π - (- arcsin (\frac{1 + 5}{4}))

使用法则

- (- a) = a

= π + arcsin (\frac{1 + 5}{4})

- arcsin (\frac{1 + 5}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

sin (u) \leq \frac{5 - 1}{4} : - π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn

sin (u) \leq \frac{5 - 1}{4}

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{5 - 1}{4}) + 2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn

合并区间

- arcsin (\frac{1 + 5}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn and - π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn

合并重叠的区间

2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn

sin (u) \geq \frac{1}{2} : \frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn

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

化简

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

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}

化简

π - arcsin (\frac{1}{2}) : \frac{5 π}{6}

π - arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

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}{6}

化简

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}

= \frac{5 π}{6}

\frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn

合并区间

(2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn) or \frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn

合并重叠的区间

2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn

2 πn \leq u \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq u \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq u \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq u < 2 π + 2 πn

\frac{x}{2} = u

代回

2 πn \leq (\frac{x}{2}) \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq (\frac{x}{2}) \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq (\frac{x}{2}) \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq (\frac{x}{2}) < 2 π + 2 πn

2 πn \leq (\frac{x}{2}) \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq (\frac{x}{2}) \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq (\frac{x}{2}) \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq (\frac{x}{2}) < 2 π + 2 πn : x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn or \frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn or 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn or x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

2 πn \leq (\frac{x}{2}) \leq arcsin (\frac{5 - 1}{4}) + 2 πn or \frac{π}{6} + 2 πn \leq (\frac{x}{2}) \leq \frac{5 π}{6} + 2 πn or π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq (\frac{x}{2}) \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn or - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq (\frac{x}{2}) < 2 π + 2 πn

2 πn \leq \frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn : x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

2 πn \leq \frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

2 πn \leq \frac{x}{2} and \frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn

2 πn \leq \frac{x}{2} : x \geq 4 πn

2 πn \leq \frac{x}{2}

交换两边

\frac{x}{2} \geq 2 πn

在两边乘以

2

\frac{x}{2} \geq 2 πn

在两边乘以

2

\frac{2 x}{2} \geq 2 \cdot 2 πn

化简

x \geq 4 πn

x \geq 4 πn

\frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn : x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

\frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} \leq arcsin (\frac{5 - 1}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} \leq 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} \leq 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn : 2 arcsin (\frac{5 - 1}{4}) + 4 πn

2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

合并区间

x \geq 4 πn and x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

合并重叠的区间

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn

\frac{π}{6} + 2 πn \leq \frac{x}{2} \leq \frac{5 π}{6} + 2 πn : \frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

\frac{π}{6} + 2 πn \leq \frac{x}{2} : x \geq \frac{π}{3} + 4 πn

\frac{π}{6} + 2 πn \leq \frac{x}{2}

交换两边

\frac{x}{2} \geq \frac{π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} \geq \frac{π}{6} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 2}{6}

约分:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

x \geq \frac{π}{3} + 4 πn

x \geq \frac{π}{3} + 4 πn

x \geq \frac{π}{3} + 4 πn

\frac{x}{2} \leq \frac{5 π}{6} + 2 πn : x \leq \frac{5 π}{3} + 4 πn

\frac{x}{2} \leq \frac{5 π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} \leq \frac{5 π}{6} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

分式相乘:

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

= \frac{5 π 2}{6}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{6}

约分:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

x \leq \frac{5 π}{3} + 4 πn

x \leq \frac{5 π}{3} + 4 πn

x \leq \frac{5 π}{3} + 4 πn

合并区间

x \geq \frac{π}{3} + 4 πn and x \leq \frac{5 π}{3} + 4 πn

合并重叠的区间

\frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn

π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq \frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn : 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq \frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq \frac{x}{2} and \frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq \frac{x}{2} : x \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

π - arcsin (\frac{5 - 1}{4}) + 2 πn \leq \frac{x}{2}

交换两边

\frac{x}{2} \geq π - arcsin (\frac{5 - 1}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} \geq π - arcsin (\frac{5 - 1}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 π - 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn : 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

2 π - 2 arcsin (\frac{5 - 1}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

x \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn

\frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn : x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

\frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} \leq π + arcsin (\frac{1 + 5}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 π + 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 πn : 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

2 π + 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

合并区间

x \geq 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn and x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

合并重叠的区间

2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn

- arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq \frac{x}{2} < 2 π + 2 πn : x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

- arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq \frac{x}{2} < 2 π + 2 πn

若

a \leq u < b

，则

a \leq u and u < b

- arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq \frac{x}{2} and \frac{x}{2} < 2 π + 2 πn

- arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq \frac{x}{2} : x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

- arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn \leq \frac{x}{2}

交换两边

\frac{x}{2} \geq - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn

在两边乘以

2

\frac{x}{2} \geq - arcsin (\frac{1 + 5}{4}) + 2 π + 2 πn

在两边乘以

2

\frac{2 x}{2} \geq - 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 π + 2 \cdot 2 πn

化简

\frac{2 x}{2} \geq - 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 π + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 π + 2 \cdot 2 πn : - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

- 2 arcsin (\frac{1 + 5}{4}) + 2 \cdot 2 π + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

\frac{x}{2} < 2 π + 2 πn : x < 4 π + 4 πn

\frac{x}{2} < 2 π + 2 πn

在两边乘以

2

\frac{x}{2} < 2 π + 2 πn

在两边乘以

2

\frac{2 x}{2} < 2 \cdot 2 π + 2 \cdot 2 πn

化简

x < 4 π + 4 πn

x < 4 π + 4 πn

合并区间

x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn and x < 4 π + 4 πn

合并重叠的区间

x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

合并区间

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn or \frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn or 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn or x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

x \leq 2 arcsin (\frac{5 - 1}{4}) + 4 πn or \frac{π}{3} + 4 πn \leq x \leq \frac{5 π}{3} + 4 πn or 2 π - 2 arcsin (\frac{5 - 1}{4}) + 4 πn \leq x \leq 2 π + 2 arcsin (\frac{1 + 5}{4}) + 4 πn or x \geq - 2 arcsin (\frac{1 + 5}{4}) + 4 π + 4 πn

流行的例子

-1<= arccos(x^2)

- 1 \leq arccos (x^{2})

2sin(5x)<= sqrt(2)

2 sin (5 x) \leq 2

(2cos(x)-sqrt(3))>0

(2 cos (x) - 3) > 0

2sin^2(x/4)<1.5

2 sin^{2} (\frac{x}{4}) < 1.5

cos^{2} (x) < \frac{3}{4}