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受欢迎的三角函数 >

cos(4/3 x+pi/6)>= 1

解答

cos (\frac{4}{3} x + \frac{π}{6}) \geq 1

解答

x = - \frac{π}{8} + \frac{3 π}{2} n

+1

十进制

x = - 0.39269 \dots + \frac{3 π}{2} n

求解步骤

cos (\frac{4}{3} x + \frac{π}{6}) \geq 1

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (1) + 2 πn \leq (\frac{4}{3} x + \frac{π}{6}) \leq arccos (1) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (1) + 2 πn \leq \frac{4}{3} x + \frac{π}{6} and \frac{4}{3} x + \frac{π}{6} \leq arccos (1) + 2 πn

- arccos (1) + 2 πn \leq \frac{4}{3} x + \frac{π}{6} : x \geq \frac{3 πn}{2} - \frac{π}{8}

- arccos (1) + 2 πn \leq \frac{4}{3} x + \frac{π}{6}

交换两边

\frac{4}{3} x + \frac{π}{6} \geq - arccos (1) + 2 πn

化简

- arccos (1) + 2 πn : 2 πn

- arccos (1) + 2 πn

使用以下普通恒等式:

arccos (1) = 0

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}

= - 0 + 2 πn

- 0 + 2 πn = 2 πn

= 2 πn

\frac{4}{3} x + \frac{π}{6} \geq 2 πn

将

\frac{π}{6}

到右边

\frac{4}{3} x + \frac{π}{6} \geq 2 πn

两边减去

\frac{π}{6}

\frac{4}{3} x + \frac{π}{6} - \frac{π}{6} \geq 2 πn - \frac{π}{6}

化简

\frac{4}{3} x \geq 2 πn - \frac{π}{6}

\frac{4}{3} x \geq 2 πn - \frac{π}{6}

在两边乘以

3

\frac{4}{3} x \geq 2 πn - \frac{π}{6}

在两边乘以

3

3 \cdot \frac{4}{3} x \geq 3 \cdot 2 πn - 3 \cdot \frac{π}{6}

化简

3 \cdot \frac{4}{3} x \geq 3 \cdot 2 πn - 3 \cdot \frac{π}{6}

化简

3 \cdot \frac{4}{3} x : 4 x

3 \cdot \frac{4}{3} x

分式相乘:

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

= \frac{4 \cdot 3}{3} x

约分:

3

= x \cdot 4

化简

3 \cdot 2 πn - 3 \cdot \frac{π}{6} : 6 πn - \frac{π}{2}

3 \cdot 2 πn - 3 \cdot \frac{π}{6}

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

数字相乘:

3 \cdot 2 = 6

= 6 πn

3 \cdot \frac{π}{6} = \frac{π}{2}

3 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 3}{6}

约分:

3

= \frac{π}{2}

= 6 πn - \frac{π}{2}

4 x \geq 6 πn - \frac{π}{2}

4 x \geq 6 πn - \frac{π}{2}

4 x \geq 6 πn - \frac{π}{2}

两边除以

4

4 x \geq 6 πn - \frac{π}{2}

两边除以

4

\frac{4 x}{4} \geq \frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

化简

\frac{4 x}{4} \geq \frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{6 πn}{4} - \frac{\frac{π}{2}}{4} : \frac{3 πn}{2} - \frac{π}{8}

\frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

\frac{6 πn}{4} = \frac{3 πn}{2}

\frac{6 πn}{4}

约分:

2

= \frac{3 πn}{2}

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

使用分式法则:

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

= \frac{π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π}{8}

= \frac{3 πn}{2} - \frac{π}{8}

x \geq \frac{3 πn}{2} - \frac{π}{8}

x \geq \frac{3 πn}{2} - \frac{π}{8}

x \geq \frac{3 πn}{2} - \frac{π}{8}

\frac{4}{3} x + \frac{π}{6} \leq arccos (1) + 2 πn : x \leq \frac{3 πn}{2} - \frac{π}{8}

\frac{4}{3} x + \frac{π}{6} \leq arccos (1) + 2 πn

化简

arccos (1) + 2 πn : 2 πn

arccos (1) + 2 πn

使用以下普通恒等式:

arccos (1) = 0

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}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

\frac{4}{3} x + \frac{π}{6} \leq 2 πn

将

\frac{π}{6}

到右边

\frac{4}{3} x + \frac{π}{6} \leq 2 πn

两边减去

\frac{π}{6}

\frac{4}{3} x + \frac{π}{6} - \frac{π}{6} \leq 2 πn - \frac{π}{6}

化简

\frac{4}{3} x \leq 2 πn - \frac{π}{6}

\frac{4}{3} x \leq 2 πn - \frac{π}{6}

在两边乘以

3

\frac{4}{3} x \leq 2 πn - \frac{π}{6}

在两边乘以

3

3 \cdot \frac{4}{3} x \leq 3 \cdot 2 πn - 3 \cdot \frac{π}{6}

化简

3 \cdot \frac{4}{3} x \leq 3 \cdot 2 πn - 3 \cdot \frac{π}{6}

化简

3 \cdot \frac{4}{3} x : 4 x

3 \cdot \frac{4}{3} x

分式相乘:

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

= \frac{4 \cdot 3}{3} x

约分:

3

= x \cdot 4

化简

3 \cdot 2 πn - 3 \cdot \frac{π}{6} : 6 πn - \frac{π}{2}

3 \cdot 2 πn - 3 \cdot \frac{π}{6}

3 \cdot 2 πn = 6 πn

3 \cdot 2 πn

数字相乘:

3 \cdot 2 = 6

= 6 πn

3 \cdot \frac{π}{6} = \frac{π}{2}

3 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 3}{6}

约分:

3

= \frac{π}{2}

= 6 πn - \frac{π}{2}

4 x \leq 6 πn - \frac{π}{2}

4 x \leq 6 πn - \frac{π}{2}

4 x \leq 6 πn - \frac{π}{2}

两边除以

4

4 x \leq 6 πn - \frac{π}{2}

两边除以

4

\frac{4 x}{4} \leq \frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

化简

\frac{4 x}{4} \leq \frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{6 πn}{4} - \frac{\frac{π}{2}}{4} : \frac{3 πn}{2} - \frac{π}{8}

\frac{6 πn}{4} - \frac{\frac{π}{2}}{4}

\frac{6 πn}{4} = \frac{3 πn}{2}

\frac{6 πn}{4}

约分:

2

= \frac{3 πn}{2}

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

使用分式法则:

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

= \frac{π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π}{8}

= \frac{3 πn}{2} - \frac{π}{8}

x \leq \frac{3 πn}{2} - \frac{π}{8}

x \leq \frac{3 πn}{2} - \frac{π}{8}

x \leq \frac{3 πn}{2} - \frac{π}{8}

合并区间

x \geq \frac{3 πn}{2} - \frac{π}{8} and x \leq \frac{3 πn}{2} - \frac{π}{8}

合并重叠的区间

x = - \frac{π}{8} + \frac{3 π}{2} n

流行的例子

sqrt(2)-2cos(a)>= 0

2 - 2 cos (a) \geq 0

sin (x) cos (x) > 0

cos(x)+sin(x)>0

cos (x) + sin (x) > 0

sqrt(2)cos(2x)<1

2 cos (2 x) < 1

12 cos (2 x) < 0