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)>2cos(2x)-0.5

解答

cos (2 x) > 2 cos (2 x) - 0.5

解答

\frac{π}{6} + πn < x < \frac{5 π}{6} + πn

+2

间隔符号

(\frac{π}{6} + πn, \frac{5 π}{6} + πn)

十进制

0.52359 \dots + πn < x < 2.61799 \dots + πn

求解步骤

cos (2 x) > 2 cos (2 x) - 0.5

令

: u = cos (2 x)

u > 2 u - 0.5

u > 2 u - 0.5 : u < 0.5

u > 2 u - 0.5

将

2 u

para o lado esquerdo

u > 2 u - 0.5

两边减去

2 u

u - 2 u > 2 u - 0.5 - 2 u

化简

- u > - 0.5

- u > - 0.5

在两边乘以

- 1

- u > - 0.5

两边乘以 -1（不等式变号）

(- u) (- 1) < (- 0.5) (- 1)

化简

u < 0.5

u < 0.5

u < 0.5

u = cos (2 x)

代回

cos (2 x) < 0.5

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

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

arccos (0.5) + 2 πn < 2 x < 2 π - arccos (0.5) + 2 πn

若

a < u < b

，则

a < u and u < b

arccos (0.5) + 2 πn < 2 x and 2 x < 2 π - arccos (0.5) + 2 πn

arccos (0.5) + 2 πn < 2 x : x > \frac{π}{6} + πn

arccos (0.5) + 2 πn < 2 x

交换两边

2 x > arccos (0.5) + 2 πn

化简

arccos (0.5) + 2 πn : \frac{π}{3} + 2 πn

arccos (0.5) + 2 πn

arccos (0.5) = \frac{π}{3}

arccos (0.5)

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

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

arccos (\frac{1}{2})

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}

= \frac{π}{3}

= \frac{π}{3} + 2 πn

2 x > \frac{π}{3} + 2 πn

两边除以

2

2 x > \frac{π}{3} + 2 πn

两边除以

2

\frac{2 x}{2} > \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} > \frac{\frac{π}{3}}{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{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x > \frac{π}{6} + πn

x > \frac{π}{6} + πn

x > \frac{π}{6} + πn

2 x < 2 π - arccos (0.5) + 2 πn : x < \frac{5 π}{6} + πn

2 x < 2 π - arccos (0.5) + 2 πn

化简

2 π - arccos (0.5) + 2 πn : 2 π - \frac{π}{3} + 2 πn

2 π - arccos (0.5) + 2 πn

arccos (0.5) = \frac{π}{3}

arccos (0.5)

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

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{3}

arccos (\frac{1}{2})

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}

= \frac{π}{3}

= 2 π - \frac{π}{3} + 2 πn

2 x < 2 π - \frac{π}{3} + 2 πn

两边除以

2

2 x < 2 π - \frac{π}{3} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} < \frac{2 π}{2} - \frac{\frac{π}{3}}{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{2 π}{2} - \frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : π - \frac{π}{6} + πn

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

\frac{2 π}{2} = π

\frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= π - \frac{π}{6} + πn

x < π - \frac{π}{6} + πn

x < π - \frac{π}{6} + πn

化简

π - \frac{π}{6} : \frac{5 π}{6}

π - \frac{π}{6}

将项转换为分式:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{π}{6}

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{π 6 - π}{6}

同类项相加:

6 π - π = 5 π

= \frac{5 π}{6}

x < \frac{5 π}{6} + πn

x < \frac{5 π}{6} + πn

合并区间

x > \frac{π}{6} + πn and x < \frac{5 π}{6} + πn

合并重叠的区间

\frac{π}{6} + πn < x < \frac{5 π}{6} + πn

流行的例子

2 sin (2 x) - 1 \geq 0

-12cos(2x)+12sin(x)>0

- 12 cos (2 x) + 12 sin (x) > 0

(-1/5)*sin(2 pi/5 (x+1))+1<= 16/15

(- \frac{1}{5}) \cdot sin (2 \frac{π}{5} (x + 1)) + 1 \leq \frac{16}{15}

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

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

3 sin (t) \geq 0