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受欢迎的三角函数 >

2cos(x)<= 1,-pi<= x<pi

解答

2 cos (x) \leq 1, - π \leq x < π

解答

- π \leq x \leq \frac{5 π}{3} - 2 π or \frac{π}{3} \leq x < π

+2

间隔符号

[- π, \frac{5 π}{3} - 2 π] \cup [\frac{π}{3}, π)

十进制

- 3.14159 \dots \leq x \leq - 1.04719 \dots or 1.04719 \dots \leq x < 3.14159 \dots

求解步骤

2 cos (x) \leq 1, - π \leq x < π

两边除以

2

2 cos (x) \leq 1

两边除以

2

\frac{2 cos ( x )}{2} \leq \frac{1}{2}

化简

cos (x) \leq \frac{1}{2}

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

化简

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{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{π}{3}

化简

2 π - arccos (\frac{1}{2}) : \frac{5 π}{3}

2 π - arccos (\frac{1}{2})

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{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}

= 2 π - \frac{π}{3}

化简

2 π - \frac{π}{3}

将项转换为分式:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

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 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

数字相乘:

2 \cdot 3 = 6

= 6 π - π

同类项相加:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

合并区间

\frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn and - π \leq x < π

- π \leq x \leq \frac{5 π}{3} - 2 π or \frac{π}{3} \leq x < π

作图

流行的例子

90-arctan((3L)/4)>= 40

90 - arctan (\frac{3 L}{4}) \geq 40

sin (x) \leq - 3

sin (x) \leq - 1

5*sin(x)<= 5/2 ,([0,2pi])

5 \cdot sin (x) \leq \frac{5}{2}, ([0, 2 π])

tan(x)+cot(x)<1

tan (x) + cot (x) < 1