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受欢迎的三角函数 >

cos(x)<= (sqrt(2))/2

解答

cos (x) \leq \frac{2}{2}

解答

\frac{π}{4} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

+2

间隔符号

[\frac{π}{4} + 2 πn, \frac{7 π}{4} + 2 πn]

十进制

0.78539 \dots + 2 πn \leq x \leq 5.49778 \dots + 2 πn

求解步骤

cos (x) \leq \frac{2}{2}

对于

cos (x) \leq a

，若

- 1 < a < 1

，则

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

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

化简

arccos (\frac{2}{2}) : \frac{π}{4}

arccos (\frac{2}{2})

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4}

化简

2 π - arccos (\frac{2}{2}) : \frac{7 π}{4}

2 π - arccos (\frac{2}{2})

使用以下普通恒等式:

arccos (\frac{2}{2}) = \frac{π}{4}

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 - \frac{π}{4}

化简

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 = 7 π

2 π 4 - π

数字相乘:

2 \cdot 4 = 8

= 8 π - π

同类项相加:

8 π - π = 7 π

= 7 π

= \frac{7 π}{4}

= \frac{7 π}{4}

\frac{π}{4} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

流行的例子

cos(φ)<= (sqrt(2))/2

cos (φ) \leq \frac{2}{2}

2 cos (y) > 0

cos(2x)<=-(sqrt(3))/2

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

sin(x)>= (sqrt(3))/2

sin (x) \geq \frac{3}{2}

2 cos (x) - 1 < 0