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受欢迎的三角函数 >

(sin(x)-1/2)(sin(x)-7/2)<= 0

解答

(sin (x) - \frac{1}{2}) (sin (x) - \frac{7}{2}) \leq 0

解答

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

+2

间隔符号

[\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn]

十进制

0.52359 \dots + 2 πn \leq x \leq 2.61799 \dots + 2 πn

求解步骤

(sin (x) - \frac{1}{2}) (sin (x) - \frac{7}{2}) \leq 0

令

: u = sin (x)

(u - \frac{1}{2}) (u - \frac{7}{2}) \leq 0

(u - \frac{1}{2}) (u - \frac{7}{2}) \leq 0 : \frac{1}{2} \leq u \leq \frac{7}{2}

(u - \frac{1}{2}) (u - \frac{7}{2}) \leq 0

改写为标准形式

(u - \frac{1}{2}) (u - \frac{7}{2}) \leq 0

乘开

(u - \frac{1}{2}) (u - \frac{7}{2}) : u^{2} - 4 u + \frac{7}{4}

(u - \frac{1}{2}) (u - \frac{7}{2})

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = u, b = - \frac{1}{2}, c = u, d = - \frac{7}{2}

= uu + u (- \frac{7}{2}) + (- \frac{1}{2}) u + (- \frac{1}{2}) (- \frac{7}{2})

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= uu - \frac{7}{2} u - \frac{1}{2} u + \frac{1}{2} \cdot \frac{7}{2}

化简

uu - \frac{7}{2} u - \frac{1}{2} u + \frac{1}{2} \cdot \frac{7}{2} : u^{2} - 4 u + \frac{7}{4}

uu - \frac{7}{2} u - \frac{1}{2} u + \frac{1}{2} \cdot \frac{7}{2}

同类项相加:

- \frac{7}{2} u - \frac{1}{2} u = - 4 u

- \frac{7}{2} u - \frac{1}{2} u

因式分解出通项

u

= u (- \frac{7}{2} - \frac{1}{2})

- \frac{7}{2} - \frac{1}{2} = - 4

- \frac{7}{2} - \frac{1}{2}

使用法则

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

= \frac{- 7 - 1}{2}

数字相减:

- 7 - 1 = - 8

= \frac{- 8}{2}

使用分式法则:

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

= - \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= - 4

= - 4 u

= uu - 4 u + \frac{1}{2} \cdot \frac{7}{2}

uu = u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

\frac{1}{2} \cdot \frac{7}{2} = \frac{7}{4}

\frac{1}{2} \cdot \frac{7}{2}

分式相乘:

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

= \frac{1 \cdot 7}{2 \cdot 2}

整理后得

= \frac{7}{4}

= u^{2} - 4 u + \frac{7}{4}

= u^{2} - 4 u + \frac{7}{4}

u^{2} - 4 u + \frac{7}{4} \leq 0

在两边乘以

4

u^{2} \cdot 4 - 4 u \cdot 4 + \frac{7}{4} \cdot 4 \leq 0 \cdot 4

4 u^{2} - 16 u + 7 \leq 0

4 u^{2} - 16 u + 7 \leq 0

分解

4 u^{2} - 16 u + 7 : (2 u - 1) (2 u - 7)

4 u^{2} - 16 u + 7

将表达式拆分成组

4 u^{2} - 16 u + 7

定义

28

的因数:

1, 2, 4, 7, 14, 28

28

约数 (因数)

找到

28

的质因数:

2, 2, 7

28

28

除以

228 = 14 \cdot 2

= 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7

乘以

28

的质因数:

4, 14

2 \cdot 2 = 4

2 \cdot 7 = 14

4, 14

4, 14

添加质因数:

2, 7

将 1 和数字

28

自身相加

1, 28

28

的因数

1, 2, 4, 7, 14, 28

28

的负因数:

- 1, - 2, - 4, - 7, - 14, - 28

将因数乘以

- 1

得到负因数

- 1, - 2, - 4, - 7, - 14, - 28

对于每两个因数

u * v = 28

，检验是否

u + v = - 16

检验

u = 1, v = 28 : u * v = 28, u + v = 29 \Rightarrow

假检验

u = 2, v = 14 : u * v = 28, u + v = 16 \Rightarrow

假

u = - 2, v = - 14

分组为

(a x^{2} + ux) + (vx + c)

(4 u^{2} - 2 u) + (- 14 u + 7)

= (4 u^{2} - 2 u) + (- 14 u + 7)

从

4 u^{2} - 2 u

分解出因式

2 u

:

2 u (2 u - 1)

4 u^{2} - 2 u

使用指数法则:

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

u^{2} = uu

= 4 uu - 2 u

将

4

改写为

2 \cdot 2

= 2 \cdot 2 uu - 2 u

因式分解出通项

2 u

= 2 u (2 u - 1)

从

- 14 u + 7

分解出因式

- 7

:

- 7 (2 u - 1)

- 14 u + 7

将

14

改写为

7 \cdot 2

= - 7 \cdot 2 u + 7

因式分解出通项

- 7

= - 7 (2 u - 1)

= 2 u (2 u - 1) - 7 (2 u - 1)

因式分解出通项

2 u - 1

= (2 u - 1) (2 u - 7)

(2 u - 1) (2 u - 7) \leq 0

确定区间

确定

(2 u - 1) (2 u - 7)

符号

确定

2 u - 1

符号

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

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

\frac{2 u}{2} = \frac{1}{2}

化简

u = \frac{1}{2}

u = \frac{1}{2}

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

2 u - 1 < 0

将

1

到右边

2 u - 1 < 0

两边加上

1

2 u - 1 + 1 < 0 + 1

化简

2 u < 1

2 u < 1

两边除以

2

2 u < 1

两边除以

2

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

化简

u < \frac{1}{2}

u < \frac{1}{2}

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

2 u - 1 > 0

将

1

到右边

2 u - 1 > 0

两边加上

1

2 u - 1 + 1 > 0 + 1

化简

2 u > 1

2 u > 1

两边除以

2

2 u > 1

两边除以

2

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

化简

u > \frac{1}{2}

u > \frac{1}{2}

确定

2 u - 7

符号

2 u - 7 = 0 : u = \frac{7}{2}

2 u - 7 = 0

将

7

到右边

2 u - 7 = 0

两边加上

7

2 u - 7 + 7 = 0 + 7

化简

2 u = 7

2 u = 7

两边除以

2

2 u = 7

两边除以

2

\frac{2 u}{2} = \frac{7}{2}

化简

u = \frac{7}{2}

u = \frac{7}{2}

2 u - 7 < 0 : u < \frac{7}{2}

2 u - 7 < 0

将

7

到右边

2 u - 7 < 0

两边加上

7

2 u - 7 + 7 < 0 + 7

化简

2 u < 7

2 u < 7

两边除以

2

2 u < 7

两边除以

2

\frac{2 u}{2} < \frac{7}{2}

化简

u < \frac{7}{2}

u < \frac{7}{2}

2 u - 7 > 0 : u > \frac{7}{2}

2 u - 7 > 0

将

7

到右边

2 u - 7 > 0

两边加上

7

2 u - 7 + 7 > 0 + 7

化简

2 u > 7

2 u > 7

两边除以

2

2 u > 7

两边除以

2

\frac{2 u}{2} > \frac{7}{2}

化简

u > \frac{7}{2}

u > \frac{7}{2}

总结如下表：

2 u - 1 2 u - 7 (2 u - 1) (2 u - 7) u < \frac{1}{2} - - + u = \frac{1}{2} 0 - 0 \frac{1}{2} < u < \frac{7}{2} + - - u = \frac{7}{2} + 00 u > \frac{7}{2} + + +

确定满足所需条件的区间：

\leq 0

u = \frac{1}{2} or \frac{1}{2} < u < \frac{7}{2} or u = \frac{7}{2}

合并重叠的区间

\frac{1}{2} \leq u < \frac{7}{2} or u = \frac{7}{2}

两个区间的并集是指存在于任一区间的数的集合

u = \frac{1}{2}

or

\frac{1}{2} < u < \frac{7}{2}

\frac{1}{2} \leq u < \frac{7}{2}

两个区间的并集是指存在于任一区间的数的集合

\frac{1}{2} \leq u < \frac{7}{2}

or

u = \frac{7}{2}

\frac{1}{2} \leq u \leq \frac{7}{2}

\frac{1}{2} \leq u \leq \frac{7}{2}

\frac{1}{2} \leq u \leq \frac{7}{2}

\frac{1}{2} \leq u \leq \frac{7}{2}

u = sin (x)

代回

\frac{1}{2} \leq sin (x) \leq \frac{7}{2}

若

a \leq u \leq b

，则

a \leq u and u \leq b

\frac{1}{2} \leq sin (x) and sin (x) \leq \frac{7}{2}

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

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

交换两边

sin (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 x \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 x \leq \frac{5 π}{6} + 2 πn

sin (x) \leq \frac{7}{2} :

对所有

x \in R

为真

sin (x) \leq \frac{7}{2}

sin (x)

的值域:

- 1 \leq sin (x) \leq 1

函数值域定义

基本

sin

函数的值域为

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) \leq \frac{7}{2} and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

令

y = sin (x)

合并区间

y \leq \frac{7}{2} and - 1 \leq y \leq 1

合并重叠的区间

y \leq \frac{7}{2} and - 1 \leq y \leq 1

两个区间的交集是指同时存在于这两个区间的数的集合

y \leq \frac{7}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

对所有 x 为真

对所有 x \in R 为真

合并区间

\frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn and 对所有 x \in R 为真

合并重叠的区间

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

流行的例子

sin (x) - 2 \leq - \frac{5}{2}

tan(1/x)<= tan(1/(x+1))

tan (\frac{1}{x}) \leq tan (\frac{1}{x + 1})

cos(x^4)+sin(x^4)>= 0.5

cos (x^{4}) + sin (x^{4}) \geq 0.5

1 - tan (x) < 2

2 sin (x) - 1 \geq - 3