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受欢迎的三角函数 >

cos(x-1)>= 0

解答

cos (x - 1) \geq 0

解答

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

+2

间隔符号

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

十进制

- 0.57079 \dots + 2 πn \leq x \leq 2.57079 \dots + 2 πn

求解步骤

cos (x - 1) \geq 0

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (0) + 2 πn \leq (x - 1) \leq arccos (0) + 2 πn

若

a \leq u \leq b

，则

a \leq u and u \leq b

- arccos (0) + 2 πn \leq x - 1 and x - 1 \leq arccos (0) + 2 πn

- arccos (0) + 2 πn \leq x - 1 : x \geq \frac{- π + 2}{2} + 2 πn

- arccos (0) + 2 πn \leq x - 1

交换两边

x - 1 \geq - arccos (0) + 2 πn

化简

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{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}{2} + 2 πn

x - 1 \geq - \frac{π}{2} + 2 πn

将

1

到右边

x - 1 \geq - \frac{π}{2} + 2 πn

两边加上

1

x - 1 + 1 \geq - \frac{π}{2} + 2 πn + 1

化简

x \geq - \frac{π}{2} + 2 πn + 1

x \geq - \frac{π}{2} + 2 πn + 1

化简

- \frac{π}{2} + 1 : \frac{- π + 2}{2}

- \frac{π}{2} + 1

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= - \frac{π}{2} + \frac{1 \cdot 2}{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{- π + 1 \cdot 2}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{- π + 2}{2}

x \geq \frac{- π + 2}{2} + 2 πn

x - 1 \leq arccos (0) + 2 πn : x \leq \frac{π + 2}{2} + 2 πn

x - 1 \leq arccos (0) + 2 πn

化简

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{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}{2} + 2 πn

x - 1 \leq \frac{π}{2} + 2 πn

将

1

到右边

x - 1 \leq \frac{π}{2} + 2 πn

两边加上

1

x - 1 + 1 \leq \frac{π}{2} + 2 πn + 1

化简

x \leq \frac{π}{2} + 2 πn + 1

x \leq \frac{π}{2} + 2 πn + 1

化简

\frac{π}{2} + 1 : \frac{π + 2}{2}

\frac{π}{2} + 1

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= \frac{π}{2} + \frac{1 \cdot 2}{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{π + 1 \cdot 2}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{π + 2}{2}

x \leq \frac{π + 2}{2} + 2 πn

合并区间

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

合并重叠的区间

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

流行的例子

sin(θ)<0,sec(θ)>0

sin (θ) < 0, sec (θ) > 0

-pi/(12)sin^2(pi/(12)t)<0

- \frac{π}{12} sin^{2} (\frac{π}{12} t) < 0

12 cos (2 x) > 0

2 cos (x) - 1 \geq 0

solvefor t,(rcos(t))/r r>0

so l v e f or t, \frac{r cos ( t )}{r} r > 0