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受欢迎的三角函数 >

(cos(x)-1)(cos(x)+1/2)>= 0

解答

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

解答

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

+2

间隔符号

[\frac{2 π}{3} + 2 πn, \frac{4 π}{3} + 2 πn]

十进制

2.09439 \dots + 2 πn \leq x \leq 4.18879 \dots + 2 πn

求解步骤

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

令

: u = cos (x)

(u - 1) (u + \frac{1}{2}) \geq 0

(u - 1) (u + \frac{1}{2}) \geq 0 : u \leq - \frac{1}{2} or u \geq 1

(u - 1) (u + \frac{1}{2}) \geq 0

改写为标准形式

(u - 1) (u + \frac{1}{2}) \geq 0

乘开

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

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

使用 FOIL 方法:

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

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

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

同类项相加:

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

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

因式分解出通项

u

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

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

\frac{1}{2} - 1

将项转换为分式:

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

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

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{1 - 1 \cdot 2}{2}

1 - 1 \cdot 2 = - 1

1 - 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= 1 - 2

数字相减:

1 - 2 = - 1

= - 1

= \frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

= - \frac{1}{2} u

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

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

1 \cdot \frac{1}{2}

乘以:

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

= \frac{1}{2}

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

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

u^{2} - \frac{1}{2} u - \frac{1}{2} \geq 0

在两边乘以

2

u^{2} \cdot 2 - \frac{1}{2} u \cdot 2 - \frac{1}{2} \cdot 2 \geq 0 \cdot 2

2 u^{2} - u - 1 \geq 0

2 u^{2} - u - 1 \geq 0

分解

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

2 u^{2} - u - 1

将表达式拆分成组

2 u^{2} - u - 1

定义

2

的因数:

1, 2

2

约数 (因数)

找到

2

的质因数:

2

2

2

是质数，因此无法因数分解

= 2

加 1

1

2

的因数

1, 2

2

的负因数:

- 1, - 2

将因数乘以

- 1

得到负因数

- 1, - 2

对于每两个因数

u * v = - 2

，检验是否

u + v = - 1

检验

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

真检验

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

假

u = 1, v = - 2

分组为

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

(2 u^{2} + u) + (- 2 u - 1)

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

从

2 u^{2} + u

分解出因式

u

:

u (2 u + 1)

2 u^{2} + u

使用指数法则:

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

u^{2} = uu

= 2 uu + u

因式分解出通项

u

= u (2 u + 1)

从

- 2 u - 1

分解出因式

- 1

:

- (2 u + 1)

- 2 u - 1

因式分解出通项

- 1

= - (2 u + 1)

= u (2 u + 1) - (2 u + 1)

因式分解出通项

2 u + 1

= (2 u + 1) (u - 1)

(2 u + 1) (u - 1) \geq 0

确定区间

确定

(2 u + 1) (u - 1)

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

两边减去

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

两边减去

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

两边减去

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

u - 1

符号

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

u - 1 < 0 : u < 1

u - 1 < 0

将

1

到右边

u - 1 < 0

两边加上

1

u - 1 + 1 < 0 + 1

化简

u < 1

u < 1

u - 1 > 0 : u > 1

u - 1 > 0

将

1

到右边

u - 1 > 0

两边加上

1

u - 1 + 1 > 0 + 1

化简

u > 1

u > 1

总结如下表：

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

确定满足所需条件的区间：

\geq 0

u < - \frac{1}{2} or u = - \frac{1}{2} or u = 1 or u > 1

合并重叠的区间

u \leq - \frac{1}{2} or u = 1 or u > 1

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

u = - \frac{1}{2}

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

两个区间的并集是指存在于任一区间的数的集合

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

or

u = 1

u \leq - \frac{1}{2} or u = 1

两个区间的并集是指存在于任一区间的数的集合

u \leq - \frac{1}{2} or u = 1

or

u > 1

u \leq - \frac{1}{2} or u \geq 1

u \leq - \frac{1}{2} or u \geq 1

u \leq - \frac{1}{2} or u \geq 1

u \leq - \frac{1}{2} or u \geq 1

u = cos (x)

代回

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

cos (x) \leq - \frac{1}{2} : \frac{2 π}{3} + 2 πn \leq x \leq \frac{4 π}{3} + 2 πn

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

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

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

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

化简

arccos (- \frac{1}{2}) : \frac{2 π}{3}

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

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

化简

2 π - arccos (- \frac{1}{2}) : \frac{4 π}{3}

2 π - arccos (- \frac{1}{2})

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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}

= 2 π - \frac{2 π}{3}

化简

2 π - \frac{2 π}{3}

将项转换为分式:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{2 π}{3}

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 π 3 - 2 π}{3}

2 π 3 - 2 π = 4 π

2 π 3 - 2 π

数字相乘:

2 \cdot 3 = 6

= 6 π - 2 π

同类项相加:

6 π - 2 π = 4 π

= 4 π

= \frac{4 π}{3}

= \frac{4 π}{3}

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

cos (x) \geq 1 : x \in R

无解

cos (x) \geq 1

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (1) + 2 πn \leq x \leq arccos (1) + 2 πn

化简

- arccos (1) : 0

- arccos (1)

使用以下普通恒等式:

arccos (1) = 0

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}

= - 0

= 0

化简

arccos (1) : 0

arccos (1)

使用以下普通恒等式:

arccos (1) = 0

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}

= 0

0 + 2 πn \leq x \leq 0 + 2 πn

化简

x \in R 无解

合并区间

\frac{2 π}{3} + 2 πn \leq x \leq \frac{4 π}{3} + 2 πn or 对所有 x \in R 为假

合并重叠的区间

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

流行的例子

tan (x) < \frac{π}{2}

cos (x - 1) \geq 0

sin(θ)<0,sec(θ)>0

sin (θ) < 0, sec (θ) > 0

-pi/(12)sin^2(pi/(12)t)<0

- \frac{π}{12} sin^{2} (\frac{π}{12} t) < 0

12 cos (2 x) > 0