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受欢迎的三角函数 >

5sin^2(x)>=-2cos(x)

解答

5 sin^{2} (x) \geq - 2 cos (x)

解答

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

+2

间隔符号

[- arccos (\frac{- 26 + 1}{5}) + 2 πn, arccos (\frac{- 26 + 1}{5}) + 2 πn]

十进制

- 2.53186 \dots + 2 πn \leq x \leq 2.53186 \dots + 2 πn

求解步骤

5 sin^{2} (x) \geq - 2 cos (x)

将

2 cos (x)

para o lado esquerdo

5 sin^{2} (x) \geq - 2 cos (x)

两边加上

2 cos (x)

5 sin^{2} (x) + 2 cos (x) \geq - 2 cos (x) + 2 cos (x)

5 sin^{2} (x) + 2 cos (x) \geq 0

5 sin^{2} (x) + 2 cos (x) \geq 0

利用以下特性:

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

因此

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

5 (1 - cos^{2} (x)) + 2 cos (x) \geq 0

令

: u = cos (x)

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

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

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

因式分解

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

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

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

5 (1 - u^{2})

分解

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

- u^{2} + 1

因式分解出通项

- 1

= - (u^{2} - 1)

分解

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

u^{2} - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (u + 1) (u - 1)

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

= - 5 (u + 1) (u - 1)

= - 5 (u + 1) (u - 1) + 2 u

乘开

- 5 (u + 1) (u - 1) + 2 u : - 5 u^{2} + 5 + 2 u

- 5 (u + 1) (u - 1) + 2 u

乘开

- 5 (u + 1) (u - 1) : - 5 u^{2} + 5

乘开

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

(u + 1) (u - 1)

使用平方差公式:

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

a = u, b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= u^{2} - 1

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

乘开

- 5 (u^{2} - 1) : - 5 u^{2} + 5

- 5 (u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = - 5, b = u^{2}, c = 1

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

使用加减运算法则

- (- a) = a

= - 5 u^{2} + 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= - 5 u^{2} + 5

= - 5 u^{2} + 5

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

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

分解

- 5 u^{2} + 2 u + 5 : - (5 u^{2} - 2 u - 5)

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

因式分解出通项

- 1

= - (5 u^{2} - 2 u - 5)

= - (5 u^{2} - 2 u - 5)

对

- (5 u^{2} - 2 u - 5)

配方:

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

- (5 u^{2} - 2 u - 5)

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

- 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

流行的例子

sin (x) > - 1

- cos (x) > 0

tan (x) \geq 3

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

sin(3x)cos(3x)-1/4 >= 0

sin (3 x) cos (3 x) - \frac{1}{4} \geq 0