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受欢迎的三角函数 >

5-5cos^2(x)+2cos(x)>= 0,0<= x<2pi

解答

5 - 5 cos^{2} (x) + 2 cos (x) \geq 0, 0 \leq x < 2 π

解答

0 \leq x \leq arccos (\frac{1 - 26}{5}) or 2 π - arccos (\frac{1 - 26}{5}) \leq x < 2 π

+2

间隔符号

[0, arccos (\frac{1 - 26}{5})] \cup [2 π - arccos (\frac{1 - 26}{5}), 2 π)

十进制

0 \leq x \leq 2.53186 \dots or 3.75132 \dots \leq x < 6.28318 \dots

求解步骤

5 - 5 cos^{2} (x) + 2 cos (x) \geq 0, 0 \leq x < 2 π

令

: u = cos (x)

5 - 5 u^{2} + 2 u \geq 0

5 - 5 u^{2} + 2 u \geq 0 : \frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

5 - 5 u^{2} + 2 u \geq 0

对

5 - 5 u^{2} + 2 u

配方:

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5}

5 - 5 u^{2} + 2 u

改写成标准形式

a x^{2} + b x + c

- 5 u^{2} + 2 u + 5

将

- 5 u^{2} + 2 u + 5

改写为如下形式:

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

因式分解

- 5

- 5 (u^{2} - \frac{2 u}{5} - 1)

2 a = - \frac{2}{5} : a = - \frac{1}{5}

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

两边除以

2

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

两边除以

2

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

化简

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

化简

\frac{2 a}{2} : a

\frac{2 a}{2}

数字相除:

\frac{2}{2} = 1

= a

化简

\frac{- \frac{2}{5}}{2} : - \frac{1}{5}

\frac{- \frac{2}{5}}{2}

使用分式法则:

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

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

使用分式法则:

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

\frac{\frac{2}{5}}{2} = \frac{2}{5 \cdot 2}

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

数字相乘:

5 \cdot 2 = 10

= - \frac{2}{10}

约分:

2

= - \frac{1}{5}

a = - \frac{1}{5}

a = - \frac{1}{5}

a = - \frac{1}{5}

加和减

(- \frac{1}{5})^{2}

- 5 (u^{2} - \frac{2 u}{5} - 1 + (- \frac{1}{5})^{2} - (- \frac{1}{5})^{2})

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

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

- 5 ((u - \frac{1}{5})^{2} - 1 - (- \frac{1}{5})^{2})

化简

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5}

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} \geq 0

将

\frac{26}{5}

到右边

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} \geq 0

两边减去

\frac{26}{5}

- 5 (u - \frac{1}{5})^{2} + \frac{26}{5} - \frac{26}{5} \geq 0 - \frac{26}{5}

化简

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

在两边乘以

- 1

- 5 (u - \frac{1}{5})^{2} \geq - \frac{26}{5}

两边乘以 -1（不等式变号）

(- 5 (u - \frac{1}{5})^{2}) (- 1) \leq (- \frac{26}{5}) (- 1)

化简

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

两边除以

5

5 (u - \frac{1}{5})^{2} \leq \frac{26}{5}

两边除以

5

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} \leq \frac{\frac{26}{5}}{5}

化简

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} \leq \frac{\frac{26}{5}}{5}

化简

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5} : (u - \frac{1}{5})^{2}

\frac{5 ( u - \frac{1}{5} ) ^{2}}{5}

数字相除:

\frac{5}{5} = 1

= (u - \frac{1}{5})^{2}

化简

\frac{\frac{26}{5}}{5} : \frac{26}{25}

\frac{\frac{26}{5}}{5}

使用分式法则:

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

= \frac{26}{5 \cdot 5}

数字相乘:

5 \cdot 5 = 25

= \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

(u - \frac{1}{5})^{2} \leq \frac{26}{25}

对于

u^{n} \leq a

，若

n

为偶数，则

- n a \leq u \leq n a

- \frac{26}{25} \leq u - \frac{1}{5} \leq \frac{26}{25}

若

a \leq u \leq b

，则

a \leq u and u \leq b

- \frac{26}{25} \leq u - \frac{1}{5} and u - \frac{1}{5} \leq \frac{26}{25}

- \frac{26}{25} \leq u - \frac{1}{5} : u \geq \frac{- 26 + 1}{5}

- \frac{26}{25} \leq u - \frac{1}{5}

交换两边

u - \frac{1}{5} \geq - \frac{26}{25}

化简

\frac{26}{25} : \frac{26}{5}

\frac{26}{25}

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{26}{25}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{26}{5}

u - \frac{1}{5} \geq - \frac{26}{5}

将

\frac{1}{5}

到右边

u - \frac{1}{5} \geq - \frac{26}{5}

两边加上

\frac{1}{5}

u - \frac{1}{5} + \frac{1}{5} \geq - \frac{26}{5} + \frac{1}{5}

化简

u - \frac{1}{5} + \frac{1}{5} \geq - \frac{26}{5} + \frac{1}{5}

化简

u - \frac{1}{5} + \frac{1}{5} : u

u - \frac{1}{5} + \frac{1}{5}

同类项相加:

- \frac{1}{5} + \frac{1}{5} \geq 0

= u

化简

- \frac{26}{5} + \frac{1}{5} : \frac{- 26 + 1}{5}

- \frac{26}{5} + \frac{1}{5}

使用法则

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

= \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u \geq \frac{- 26 + 1}{5}

u - \frac{1}{5} \leq \frac{26}{25} : u \leq \frac{26 + 1}{5}

u - \frac{1}{5} \leq \frac{26}{25}

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

u - \frac{1}{5} \leq \frac{26}{25}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u - \frac{1}{5} \leq \frac{26}{5}

将

\frac{1}{5}

到右边

u - \frac{1}{5} \leq \frac{26}{5}

两边加上

\frac{1}{5}

u - \frac{1}{5} + \frac{1}{5} \leq \frac{26}{5} + \frac{1}{5}

化简

u - \frac{1}{5} + \frac{1}{5} \leq \frac{26}{5} + \frac{1}{5}

化简

u - \frac{1}{5} + \frac{1}{5} : u

u - \frac{1}{5} + \frac{1}{5}

同类项相加:

- \frac{1}{5} + \frac{1}{5} \leq 0

= u

化简

\frac{26}{5} + \frac{1}{5} : \frac{26 + 1}{5}

\frac{26}{5} + \frac{1}{5}

使用法则

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

= \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

u \leq \frac{26 + 1}{5}

合并区间

u \geq \frac{- 26 + 1}{5} and u \leq \frac{26 + 1}{5}

合并重叠的区间

u \geq \frac{- 26 + 1}{5} and u \leq \frac{26 + 1}{5}

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 \geq \frac{- 26 + 1}{5}

and

u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq u \leq \frac{26 + 1}{5}

u = cos (x)

代回

\frac{- 26 + 1}{5} \leq cos (x) \leq \frac{26 + 1}{5}

若

a \leq u \leq b

，则

a \leq u and u \leq b

\frac{- 26 + 1}{5} \leq cos (x) and cos (x) \leq \frac{26 + 1}{5}

\frac{- 26 + 1}{5} \leq cos (x) : - arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

\frac{- 26 + 1}{5} \leq cos (x)

交换两边

cos (x) \geq \frac{- 26 + 1}{5}

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

cos (x) \leq \frac{26 + 1}{5} :

对所有

x \in R

为真

cos (x) \leq \frac{26 + 1}{5}

cos (x)

的值域:

- 1 \leq cos (x) \leq 1

函数值域定义

基本

cos

函数的值域为

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) \leq \frac{26 + 1}{5} and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

令

y = cos (x)

合并区间

y \leq \frac{26 + 1}{5} and - 1 \leq y \leq 1

合并重叠的区间

y \leq \frac{26 + 1}{5} 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{26 + 1}{5}

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

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn and 对所有 x \in R 为真

合并重叠的区间

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn

合并区间

- arccos (\frac{- 26 + 1}{5}) + 2 πn \leq x \leq arccos (\frac{- 26 + 1}{5}) + 2 πn and 0 \leq x < 2 π

0 \leq x \leq arccos (\frac{1 - 26}{5}) or 2 π - arccos (\frac{1 - 26}{5}) \leq x < 2 π

作图

流行的例子

2+cos^2(x)+sin(x)>0

2 + cos^{2} (x) + sin (x) > 0

2 sin^{2} (x) - 1 > 0

10<5sin(pi/6 (x+2))+6

10 < 5 sin (\frac{π}{6} (x + 2)) + 6

arctan(1-|x|)<= 1/2

arctan (1 - ∣ x ∣) \leq \frac{1}{2}

sin(t)< 1/3 ln(1)

sin (t) < \frac{1}{3} ln (1)