zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

-cos(2x)<= (sqrt(3))/2

解答

- cos (2 x) \leq \frac{3}{2}

解答

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

+2

间隔符号

[- \frac{5 π}{12} + πn, \frac{5 π}{12} + πn]

十进制

- 1.30899 \dots + πn \leq x \leq 1.30899 \dots + πn

求解步骤

- cos (2 x) \leq \frac{3}{2}

在两边乘以

- 1

- cos (2 x) \leq \frac{3}{2}

两边乘以 -1（不等式变号）

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

化简

cos (2 x) \geq - \frac{3}{2}

cos (2 x) \geq - \frac{3}{2}

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

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

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x and 2 x \leq arccos (- \frac{3}{2}) + 2 πn

- arccos (- \frac{3}{2}) + 2 πn \leq 2 x : x \geq - \frac{5 π}{12} + πn

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

交换两边

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

化简

- arccos (- \frac{3}{2}) + 2 πn : - \frac{5 π}{6} + 2 πn

- arccos (- \frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6} + 2 πn

2 x \geq - \frac{5 π}{6} + 2 πn

两边除以

2

2 x \geq - \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} \geq - \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \geq - \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : - \frac{5 π}{12} + πn

- \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

x \geq - \frac{5 π}{12} + πn

2 x \leq arccos (- \frac{3}{2}) + 2 πn : x \leq \frac{5 π}{12} + πn

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

化简

arccos (- \frac{3}{2}) + 2 πn : \frac{5 π}{6} + 2 πn

arccos (- \frac{3}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6} + 2 πn

2 x \leq \frac{5 π}{6} + 2 πn

两边除以

2

2 x \leq \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} \leq \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} \leq \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

x \leq \frac{5 π}{12} + πn

合并区间

x \geq - \frac{5 π}{12} + πn and x \leq \frac{5 π}{12} + πn

合并重叠的区间

- \frac{5 π}{12} + πn \leq x \leq \frac{5 π}{12} + πn

流行的例子

sin(x)<0,sec(x)>0

sin (x) < 0, sec (x) > 0

2sin(x)+3((sin(2x))/(2sin(x)))<0

2 sin (x) + 3 (\frac{sin ( 2 x )}{2 sin ( x )}) < 0

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

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

cos(θ)>0,sin(θ)>0

cos (θ) > 0, sin (θ) > 0

tan^2(x)>= sqrt(3)tan(x)

tan^{2} (x) \geq 3 tan (x)