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^2(x)-1/4 <= 0

解答

cos^{2} (x) - \frac{1}{4} \leq 0

解答

\frac{π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

+2

间隔符号

[\frac{π}{3} + 2 πn, \frac{2 π}{3} + 2 πn] \cup [\frac{4 π}{3} + 2 πn, \frac{5 π}{3} + 2 πn]

十进制

1.04719 \dots + 2 πn \leq x \leq 2.09439 \dots + 2 πn or 4.18879 \dots + 2 πn \leq x \leq 5.23598 \dots + 2 πn

求解步骤

cos^{2} (x) - \frac{1}{4} \leq 0

将

\frac{1}{4}

到右边

cos^{2} (x) - \frac{1}{4} \leq 0

两边加上

\frac{1}{4}

cos^{2} (x) - \frac{1}{4} + \frac{1}{4} \leq 0 + \frac{1}{4}

化简

cos^{2} (x) \leq \frac{1}{4}

cos^{2} (x) \leq \frac{1}{4}

对于

u^{n} \leq a

，若

n

为偶数，则

- n a \leq u \leq n a

- \frac{1}{4} \leq cos (x) \leq \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{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{1}{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{1}{2}

使用法则

1 = 1

= \frac{1}{2}

- \frac{1}{2} \leq cos (x) \leq \frac{1}{2}

若

a \leq u \leq b

，则

a \leq u and u \leq b

- \frac{1}{2} \leq cos (x) and cos (x) \leq \frac{1}{2}

- \frac{1}{2} \leq cos (x) : - \frac{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn

- \frac{1}{2} \leq cos (x)

交换两边

cos (x) \geq - \frac{1}{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{1}{2}) + 2 πn \leq x \leq arccos (- \frac{1}{2}) + 2 πn

化简

- arccos (- \frac{1}{2}) : - \frac{2 π}{3}

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

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

化简

arccos (- \frac{1}{2}) : \frac{2 π}{3}

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

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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{2 π}{3}

- \frac{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn

cos (x) \leq \frac{1}{2} : \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

cos (x) \leq \frac{1}{2}

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

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

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

化简

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

化简

2 π - arccos (\frac{1}{2}) : \frac{5 π}{3}

2 π - arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

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}

= 2 π - \frac{π}{3}

化简

2 π - \frac{π}{3}

将项转换为分式:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

因为分母相等，所以合并分式:

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

= \frac{2 π 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

数字相乘:

2 \cdot 3 = 6

= 6 π - π

同类项相加:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

合并区间

- \frac{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn and \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

合并重叠的区间

\frac{π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

流行的例子

-sin(θ)+cos(θ)-sqrt(10)<0

- sin (θ) + cos (θ) - 10 < 0

cos(x)+sqrt(3)sin(x)>= sqrt(2)

cos (x) + 3 sin (x) \geq 2

sin(0.5pix)<0.6

sin (0.5 π x) < 0.6

2 + sin (x) \geq 0

cos^2(θ)-1>= 0

cos^{2} (θ) - 1 \geq 0