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受欢迎的三角函数 >

2cos(x)>cos(2x)

解答

2 cos (x) > cos (2 x)

解答

- arccos (\frac{- 3 + 1}{2}) + 2 πn < x < arccos (\frac{- 3 + 1}{2}) + 2 πn

+2

间隔符号

(- arccos (\frac{- 3 + 1}{2}) + 2 πn, arccos (\frac{- 3 + 1}{2}) + 2 πn)

十进制

- 1.94553 \dots + 2 πn < x < 1.94553 \dots + 2 πn

求解步骤

2 cos (x) > cos (2 x)

将

cos (2 x)

para o lado esquerdo

2 cos (x) > cos (2 x)

两边减去

cos (2 x)

2 cos (x) - cos (2 x) > cos (2 x) - cos (2 x)

2 cos (x) - cos (2 x) > 0

2 cos (x) - cos (2 x) > 0

利用以下特性:

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

- (- 1 + 2 cos^{2} (x)) + 2 cos (x) > 0

化简

1 - 2 cos^{2} (x) + 2 cos (x) > 0

令

: u = cos (x)

1 - 2 u^{2} + 2 u > 0

1 - 2 u^{2} + 2 u > 0 : \frac{- 3 + 1}{2} < u < \frac{3 + 1}{2}

1 - 2 u^{2} + 2 u > 0

对

1 - 2 u^{2} + 2 u

配方:

- 2 (u - \frac{1}{2})^{2} + \frac{3}{2}

1 - 2 u^{2} + 2 u

改写成标准形式

a x^{2} + b x + c

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

将

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

改写为如下形式:

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

因式分解

- 2

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

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}

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

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

化简

- 2 (u - \frac{1}{2})^{2} + \frac{3}{2}

- 2 (u - \frac{1}{2})^{2} + \frac{3}{2} > 0

将

\frac{3}{2}

到右边

- 2 (u - \frac{1}{2})^{2} + \frac{3}{2} > 0

两边减去

\frac{3}{2}

- 2 (u - \frac{1}{2})^{2} + \frac{3}{2} - \frac{3}{2} > 0 - \frac{3}{2}

化简

- 2 (u - \frac{1}{2})^{2} > - \frac{3}{2}

- 2 (u - \frac{1}{2})^{2} > - \frac{3}{2}

在两边乘以

- 1

- 2 (u - \frac{1}{2})^{2} > - \frac{3}{2}

两边乘以 -1（不等式变号）

(- 2 (u - \frac{1}{2})^{2}) (- 1) < (- \frac{3}{2}) (- 1)

化简

2 (u - \frac{1}{2})^{2} < \frac{3}{2}

2 (u - \frac{1}{2})^{2} < \frac{3}{2}

两边除以

2

2 (u - \frac{1}{2})^{2} < \frac{3}{2}

两边除以

2

\frac{2 ( u - \frac{1}{2} ) ^{2}}{2} < \frac{\frac{3}{2}}{2}

化简

\frac{2 ( u - \frac{1}{2} ) ^{2}}{2} < \frac{\frac{3}{2}}{2}

化简

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

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

数字相除:

\frac{2}{2} = 1

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

化简

\frac{\frac{3}{2}}{2} : \frac{3}{4}

\frac{\frac{3}{2}}{2}

使用分式法则:

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

= \frac{3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3}{4}

(u - \frac{1}{2})^{2} < \frac{3}{4}

(u - \frac{1}{2})^{2} < \frac{3}{4}

(u - \frac{1}{2})^{2} < \frac{3}{4}

对于

u^{n} < a

，若

n

为偶数，则

- n a < u < n a

- \frac{3}{4} < u - \frac{1}{2} < \frac{3}{4}

若

a < u < b

，则

a < u and u < b

- \frac{3}{4} < u - \frac{1}{2} and u - \frac{1}{2} < \frac{3}{4}

- \frac{3}{4} < u - \frac{1}{2} : u > \frac{- 3 + 1}{2}

- \frac{3}{4} < u - \frac{1}{2}

交换两边

u - \frac{1}{2} > - \frac{3}{4}

化简

\frac{3}{4} : \frac{3}{2}

\frac{3}{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{3}{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{3}{2}

u - \frac{1}{2} > - \frac{3}{2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} > - \frac{3}{2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} > - \frac{3}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} > - \frac{3}{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} > 0

= u

化简

- \frac{3}{2} + \frac{1}{2} : \frac{- 3 + 1}{2}

- \frac{3}{2} + \frac{1}{2}

使用法则

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

= \frac{- 3 + 1}{2}

u > \frac{- 3 + 1}{2}

u > \frac{- 3 + 1}{2}

u > \frac{- 3 + 1}{2}

u - \frac{1}{2} < \frac{3}{4} : u < \frac{3 + 1}{2}

u - \frac{1}{2} < \frac{3}{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} < \frac{3}{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} < \frac{3}{2}

将

\frac{1}{2}

到右边

u - \frac{1}{2} < \frac{3}{2}

两边加上

\frac{1}{2}

u - \frac{1}{2} + \frac{1}{2} < \frac{3}{2} + \frac{1}{2}

化简

u - \frac{1}{2} + \frac{1}{2} < \frac{3}{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} < 0

= u

化简

\frac{3}{2} + \frac{1}{2} : \frac{3 + 1}{2}

\frac{3}{2} + \frac{1}{2}

使用法则

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

= \frac{3 + 1}{2}

u < \frac{3 + 1}{2}

u < \frac{3 + 1}{2}

u < \frac{3 + 1}{2}

合并区间

u > \frac{- 3 + 1}{2} and u < \frac{3 + 1}{2}

合并重叠的区间

u > \frac{- 3 + 1}{2} and u < \frac{3 + 1}{2}

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 > \frac{- 3 + 1}{2}

and

u < \frac{3 + 1}{2}

\frac{- 3 + 1}{2} < u < \frac{3 + 1}{2}

\frac{- 3 + 1}{2} < u < \frac{3 + 1}{2}

\frac{- 3 + 1}{2} < u < \frac{3 + 1}{2}

u = cos (x)

代回

\frac{- 3 + 1}{2} < cos (x) < \frac{3 + 1}{2}

若

a < u < b

，则

a < u and u < b

\frac{- 3 + 1}{2} < cos (x) and cos (x) < \frac{3 + 1}{2}

\frac{- 3 + 1}{2} < cos (x) : - arccos (\frac{- 3 + 1}{2}) + 2 πn < x < arccos (\frac{- 3 + 1}{2}) + 2 πn

\frac{- 3 + 1}{2} < cos (x)

交换两边

cos (x) > \frac{- 3 + 1}{2}

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (\frac{- 3 + 1}{2}) + 2 πn < x < arccos (\frac{- 3 + 1}{2}) + 2 πn

cos (x) < \frac{3 + 1}{2} :

对所有

x \in R

为真

cos (x) < \frac{3 + 1}{2}

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) < \frac{3 + 1}{2} and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

令

y = cos (x)

合并区间

y < \frac{3 + 1}{2} and - 1 \leq y \leq 1

合并重叠的区间

y < \frac{3 + 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 < \frac{3 + 1}{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

- arccos (\frac{- 3 + 1}{2}) + 2 πn < x < arccos (\frac{- 3 + 1}{2}) + 2 πn and 对所有 x \in R 为真

合并重叠的区间

- arccos (\frac{- 3 + 1}{2}) + 2 πn < x < arccos (\frac{- 3 + 1}{2}) + 2 πn

流行的例子

tan^{2} (x) < 1

sin(2t)<1,(0,2pi)

sin (2 t) < 1, (0, 2 π)

2sin^2(x)-3sin(x)+1<= 0

2 sin^{2} (x) - 3 sin (x) + 1 \leq 0

cos(x)-1/2 cos(2x)<0

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

cot(θ)<sqrt(3)

cot (θ) < 3