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受欢迎的三角函数 >

cos(x)+(sqrt(3))/2 <= 0

解答

cos (x) + \frac{3}{2} \leq 0

解答

\frac{5 π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

+2

间隔符号

[\frac{5 π}{6} + 2 πn, \frac{7 π}{6} + 2 πn]

十进制

2.61799 \dots + 2 πn \leq x \leq 3.66519 \dots + 2 πn

求解步骤

cos (x) + \frac{3}{2} \leq 0

将

\frac{3}{2}

到右边

cos (x) + \frac{3}{2} \leq 0

两边减去

\frac{3}{2}

cos (x) + \frac{3}{2} - \frac{3}{2} \leq 0 - \frac{3}{2}

化简

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

cos (x) \leq - \frac{3}{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{3}{2}) + 2 πn \leq x \leq 2 π - arccos (- \frac{3}{2}) + 2 πn

化简

arccos (- \frac{3}{2}) : \frac{5 π}{6}

arccos (- \frac{3}{2})

使用以下普通恒等式:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6}

化简

2 π - arccos (- \frac{3}{2}) : \frac{7 π}{6}

2 π - arccos (- \frac{3}{2})

使用以下普通恒等式:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

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{5 π}{6}

化简

2 π - \frac{5 π}{6}

将项转换为分式:

2 π = \frac{2 π 6}{6}

= \frac{2 π 6}{6} - \frac{5 π}{6}

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 6 - 5 π}{6}

2 π 6 - 5 π = 7 π

2 π 6 - 5 π

数字相乘:

2 \cdot 6 = 12

= 12 π - 5 π

同类项相加:

12 π - 5 π = 7 π

= 7 π

= \frac{7 π}{6}

= \frac{7 π}{6}

\frac{5 π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

流行的例子

(sin(x)+cos(x))>= 1/2

(sin (x) + cos (x)) \geq \frac{1}{2}

tan(x)<= sqrt(3)

tan (x) \leq 3

50sin(-pi/2 x-pi/2)>= 15

50 sin (- \frac{π}{2} x - \frac{π}{2}) \geq 15

solvefor x,sin(x)cos(2x)>0

so l v e f or x, sin (x) cos (2 x) > 0

cos (2 x) \geq 1