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受欢迎的三角函数 >

cos^2(x)<=-sin(x)

解答

cos^{2} (x) \leq - sin (x)

解答

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

+2

间隔符号

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

十进制

- 2.47535 \dots + 2 πn \leq x \leq - 0.66623 \dots + 2 πn

求解步骤

cos^{2} (x) \leq - sin (x)

将

sin (x)

para o lado esquerdo

cos^{2} (x) \leq - sin (x)

两边加上

sin (x)

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

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

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

利用以下特性:

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

因此

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

1 - sin^{2} (x) + sin (x) \leq 0

令

: u = sin (x)

1 - u^{2} + u \leq 0

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

1 - u^{2} + u \leq 0

对

1 - u^{2} + u

配方:

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

1 - u^{2} + u

改写成标准形式

a x^{2} + b x + c

- u^{2} + u + 1

将

- u^{2} + u + 1

改写为如下形式:

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

因式分解

- 1

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

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

2 a = - 1

两边除以

2

2 a = - 1

两边除以

2

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

化简

a = - \frac{1}{2}

a = - \frac{1}{2}

加和减

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

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

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

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

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

化简

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

- (u - \frac{1}{2})^{2} + \frac{5}{4} \leq 0

将

\frac{5}{4}

到右边

- (u - \frac{1}{2})^{2} + \frac{5}{4} \leq 0

两边减去

\frac{5}{4}

- (u - \frac{1}{2})^{2} + \frac{5}{4} - \frac{5}{4} \leq 0 - \frac{5}{4}

化简

- (u - \frac{1}{2})^{2} \leq - \frac{5}{4}

- (u - \frac{1}{2})^{2} \leq - \frac{5}{4}

在两边乘以

- 1

- (u - \frac{1}{2})^{2} \leq - \frac{5}{4}

两边乘以 -1（不等式变号）

(- (u - \frac{1}{2})^{2}) (- 1) \geq (- \frac{5}{4}) (- 1)

化简

(u - \frac{1}{2})^{2} \geq \frac{5}{4}

(u - \frac{1}{2})^{2} \geq \frac{5}{4}

对于

u^{n} \geq a

，若

n

为偶数，则

u \leq - n a or u \geq n a

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

u - \frac{1}{2} \leq - \frac{5}{4} : u \leq \frac{- 5 + 1}{2}

u - \frac{1}{2} \leq - \frac{5}{4}

化简

\frac{5}{4} : \frac{5}{2}

\frac{5}{4}

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{5}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{5}{2}

u - \frac{1}{2} \leq - \frac{5}{2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} \leq - \frac{5}{2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} \leq - \frac{5}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} \leq - \frac{5}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} : u

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

同类项相加:

- \frac{1}{2} + \frac{1}{2} \leq 0

= u

化简

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

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

使用法则

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

= \frac{- 5 + 1}{2}

u \leq \frac{- 5 + 1}{2}

u \leq \frac{- 5 + 1}{2}

u \leq \frac{- 5 + 1}{2}

u - \frac{1}{2} \geq \frac{5}{4} : u \geq \frac{5 + 1}{2}

u - \frac{1}{2} \geq \frac{5}{4}

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}{2} \geq \frac{5}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u - \frac{1}{2} \geq \frac{5}{2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} \geq \frac{5}{2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} \geq \frac{5}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} \geq \frac{5}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} : u

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

同类项相加:

- \frac{1}{2} + \frac{1}{2} \geq 0

= u

化简

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

\frac{5}{2} + \frac{1}{2}

使用法则

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

= \frac{5 + 1}{2}

u \geq \frac{5 + 1}{2}

u \geq \frac{5 + 1}{2}

u \geq \frac{5 + 1}{2}

合并区间

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

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

u = sin (x)

代回

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

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

sin (x) \leq \frac{- 5 + 1}{2}

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

sin (x) \geq \frac{5 + 1}{2} :

对所有

x \in R

为假

sin (x) \geq \frac{5 + 1}{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) \geq \frac{5 + 1}{2} and - 1 \leq sin (x) \leq 1 :

假

令

y = sin (x)

合并区间

y \geq \frac{5 + 1}{2} and - 1 \leq y \leq 1

合并重叠的区间

y \geq \frac{5 + 1}{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 \geq \frac{5 + 1}{2}

and

- 1 \leq y \leq 1

对所有 y \in R 为假

对所有 y \in R 为假

x \in R 无解

对所有 x \in R 为假

合并区间

- π - arcsin (\frac{- 5 + 1}{2}) + 2 πn \leq x \leq arcsin (\frac{- 5 + 1}{2}) + 2 πn or 对所有 x \in R 为假

合并重叠的区间

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

流行的例子

tan(θ)>0,sin(θ)<0

tan (θ) > 0, sin (θ) < 0

(1+2sin(x))/((cos(2x)))>0

\frac{1 + 2 sin ( x )}{( cos ( 2 x ) )} > 0

(cos^2(x)-1/2)/(tan(x)-sqrt(3))<0

\frac{cos ^{2} ( x ) - \frac{1}{2}}{tan ( x ) - 3} < 0

tan(θ)>0,cot(θ)>0

tan (θ) > 0, cot (θ) > 0

6 sin (θ) \geq 0