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受欢迎的三角函数 >

cos(x-pi/4)<1

解答

cos (x - \frac{π}{4}) < 1

解答

\frac{π}{4} + 2 πn < x < \frac{9 π}{4} + 2 πn

+2

间隔符号

(\frac{π}{4} + 2 πn, \frac{9 π}{4} + 2 πn)

十进制

0.78539 \dots + 2 πn < x < 7.06858 \dots + 2 πn

求解步骤

cos (x - \frac{π}{4}) < 1

对于

cos (x) < a

，若

- 1 < a \leq 1

，则

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

arccos (1) + 2 πn < (x - \frac{π}{4}) < 2 π - arccos (1) + 2 πn

若

a < u < b

，则

a < u and u < b

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

arccos (1) + 2 πn < x - \frac{π}{4} : x > 2 πn + \frac{π}{4}

arccos (1) + 2 πn < x - \frac{π}{4}

交换两边

x - \frac{π}{4} > 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

x - \frac{π}{4} > 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} > 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} > 2 πn + \frac{π}{4}

化简

x > 2 πn + \frac{π}{4}

x > 2 πn + \frac{π}{4}

x - \frac{π}{4} < 2 π - arccos (1) + 2 πn : x < \frac{9 π}{4} + 2 πn

x - \frac{π}{4} < 2 π - arccos (1) + 2 πn

化简

2 π - arccos (1) + 2 πn : 2 π + 2 πn

2 π - 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}

= 2 π - 0 + 2 πn

2 π - 0 + 2 πn = 2 π + 2 πn

= 2 π + 2 πn

x - \frac{π}{4} < 2 π + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} < 2 π + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} < 2 π + 2 πn + \frac{π}{4}

化简

x < 2 π + 2 πn + \frac{π}{4}

x < 2 π + 2 πn + \frac{π}{4}

化简

2 π + \frac{π}{4} : \frac{9 π}{4}

2 π + \frac{π}{4}

将项转换为分式:

2 π = \frac{2 π 4}{4}

= \frac{2 π 4}{4} + \frac{π}{4}

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 4 + π}{4}

2 π 4 + π = 9 π

2 π 4 + π

数字相乘:

2 \cdot 4 = 8

= 8 π + π

同类项相加:

8 π + π = 9 π

= 9 π

= \frac{9 π}{4}

x < \frac{9 π}{4} + 2 πn

合并区间

x > 2 πn + \frac{π}{4} and x < \frac{9 π}{4} + 2 πn

合并重叠的区间

\frac{π}{4} + 2 πn < x < \frac{9 π}{4} + 2 πn

流行的例子

cot(x/2+pi/6)>= 1/(sqrt(3))

cot (\frac{x}{2} + \frac{π}{6}) \geq \frac{1}{3}

(1-cot(x))(csc(x-pi/3)-2)<0,-pi<,x<0

(1 - cot (x)) (csc (x - \frac{π}{3}) - 2) < 0, - π <, x < 0

(cos (x))^{2} \geq 1

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

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

2cos(4x-pi/3)-1<= 0

2 cos (4 x - \frac{π}{3}) - 1 \leq 0