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受欢迎的三角函数 >

cos(x/2)>= (sqrt(2))/2

解答

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

解答

- \frac{π}{2} + 4 πn \leq x \leq \frac{π}{2} + 4 πn

+2

间隔符号

[- \frac{π}{2} + 4 πn, \frac{π}{2} + 4 πn]

十进制

- 1.57079 \dots + 4 πn \leq x \leq 1.57079 \dots + 4 πn

求解步骤

cos (\frac{x}{2}) \geq \frac{2}{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{2}{2}) + 2 πn \leq \frac{x}{2} \leq arccos (\frac{2}{2}) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

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

- arccos (\frac{2}{2}) + 2 πn \leq \frac{x}{2} : x \geq - \frac{π}{2} + 4 πn

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

交换两边

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

化简

- arccos (\frac{2}{2}) + 2 πn : - \frac{π}{4} + 2 πn

- arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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{π}{4} + 2 πn

\frac{x}{2} \geq - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} \geq - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} \geq - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} \geq - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 \cdot \frac{π}{4} + 2 \cdot 2 πn : - \frac{π}{2} + 4 πn

- 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= - \frac{π}{2} + 4 πn

x \geq - \frac{π}{2} + 4 πn

x \geq - \frac{π}{2} + 4 πn

x \geq - \frac{π}{2} + 4 πn

\frac{x}{2} \leq arccos (\frac{2}{2}) + 2 πn : x \leq \frac{π}{2} + 4 πn

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

化简

arccos (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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{π}{4} + 2 πn

\frac{x}{2} \leq \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} \leq \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} \leq 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} \leq 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{4} + 2 \cdot 2 πn : \frac{π}{2} + 4 πn

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

x \leq \frac{π}{2} + 4 πn

x \leq \frac{π}{2} + 4 πn

x \leq \frac{π}{2} + 4 πn

合并区间

x \geq - \frac{π}{2} + 4 πn and x \leq \frac{π}{2} + 4 πn

合并重叠的区间

- \frac{π}{2} + 4 πn \leq x \leq \frac{π}{2} + 4 πn

流行的例子

cos^{2} (x) > - 2

-cos(x)-4sin(2x)>0

- cos (x) - 4 sin (2 x) > 0

(1+tan(x))/(1-tan(x))>0

\frac{1 + tan ( x )}{1 - tan ( x )} > 0

0.96(cos(x))^2<= 0.83

0.96 (cos (x))^{2} \leq 0.83

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

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