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受欢迎的三角函数 >

3cos(x)+11/2 >4

解答

3 cos (x) + \frac{11}{2} > 4

解答

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

+2

间隔符号

(- \frac{2 π}{3} + 2 πn, \frac{2 π}{3} + 2 πn)

十进制

- 2.09439 \dots + 2 πn < x < 2.09439 \dots + 2 πn

求解步骤

3 cos (x) + \frac{11}{2} > 4

将

\frac{11}{2}

到右边

3 cos (x) + \frac{11}{2} > 4

两边减去

\frac{11}{2}

3 cos (x) + \frac{11}{2} - \frac{11}{2} > 4 - \frac{11}{2}

化简

3 cos (x) + \frac{11}{2} - \frac{11}{2} > 4 - \frac{11}{2}

化简

3 cos (x) + \frac{11}{2} - \frac{11}{2} : 3 cos (x)

3 cos (x) + \frac{11}{2} - \frac{11}{2}

同类项相加:

\frac{11}{2} - \frac{11}{2} > 0

= 3 cos (x)

化简

4 - \frac{11}{2} : - \frac{3}{2}

4 - \frac{11}{2}

将项转换为分式:

4 = \frac{4 \cdot 2}{2}

= \frac{4 \cdot 2}{2} - \frac{11}{2}

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{4 \cdot 2 - 11}{2}

4 \cdot 2 - 11 = - 3

4 \cdot 2 - 11

数字相乘:

4 \cdot 2 = 8

= 8 - 11

数字相减:

8 - 11 = - 3

= - 3

= \frac{- 3}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

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

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

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

两边除以

3

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

两边除以

3

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

化简

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

化简

\frac{3 cos ( x )}{3} : cos (x)

\frac{3 cos ( x )}{3}

数字相除:

\frac{3}{3} = 1

= cos (x)

化简

\frac{- \frac{3}{2}}{3} : - \frac{1}{2}

\frac{- \frac{3}{2}}{3}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{3}

使用分式法则:

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

\frac{\frac{3}{2}}{3} = \frac{3}{2 \cdot 3}

= - \frac{3}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= - \frac{3}{6}

约分:

3

= - \frac{1}{2}

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

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

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

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

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

- arccos (- \frac{1}{2}) + 2 πn < x < 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 < x < \frac{2 π}{3} + 2 πn

流行的例子

cos(x^4)+sin(x^4)<0.5

cos (x^{4}) + sin (x^{4}) < 0.5

cos((x-60)/2)>0

cos (\frac{x - 60}{2}) > 0

sin(2x-(2pi)/3)<= (sqrt(2))/2

sin (2 x - \frac{2 π}{3}) \leq \frac{2}{2}

tan(x)>tan(pi/4)

tan (x) > tan (\frac{π}{4})

(sin(x))/(cos(x))>= 2sin(x)*cos(x)

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