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受欢迎的三角函数 >

2cos^2(x)>= 1.5

解答

2 cos^{2} (x) \geq 1.5

解答

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

+2

间隔符号

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

十进制

- 0.52359 \dots + 2 πn \leq x \leq 0.52359 \dots + 2 πn or 2.61799 \dots + 2 πn \leq x \leq 3.66519 \dots + 2 πn

求解步骤

2 cos^{2} (x) \geq 1.5

两边除以

2

2 cos^{2} (x) \geq 1.5

两边除以

2

\frac{2 cos ^{2} ( x )}{2} \geq \frac{1.5}{2}

化简

cos^{2} (x) \geq 0.75

cos^{2} (x) \geq 0.75

对于

u^{n} \geq a

，若

n

为偶数，则

u \leq - n a or u \geq n a

cos (x) \leq - 0.75 or cos (x) \geq 0.75

0.75 = 0.86602 \dots

0.75

0.75 = 0.86602 \dots

= 0.86602 \dots

cos (x) \leq - 0.86602 \dots or cos (x) \geq 0.86602 \dots

cos (x) \leq - 0.86602 \dots : \frac{5 π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

cos (x) \leq - 0.86602 \dots

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 (- 0.86602 \dots) + 2 πn \leq x \leq 2 π - arccos (- 0.86602 \dots) + 2 πn

化简

arccos (- 0.86602 \dots) : \frac{5 π}{6}

arccos (- 0.86602 \dots)

使用以下普通恒等式:

arccos (- 0.86602 \dots) = \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 (- 0.86602 \dots) : \frac{7 π}{6}

2 π - arccos (- 0.86602 \dots)

使用以下普通恒等式:

arccos (- 0.86602 \dots) = \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

cos (x) \geq 0.86602 \dots : - \frac{π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

cos (x) \geq 0.86602 \dots

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (0.86602 \dots) + 2 πn \leq x \leq arccos (0.86602 \dots) + 2 πn

化简

- arccos (0.86602 \dots) : - \frac{π}{6}

- arccos (0.86602 \dots)

使用以下普通恒等式:

arccos (0.86602 \dots) = \frac{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{π}{6}

化简

arccos (0.86602 \dots) : \frac{π}{6}

arccos (0.86602 \dots)

使用以下普通恒等式:

arccos (0.86602 \dots) = \frac{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{π}{6}

- \frac{π}{6} + 2 πn \leq x \leq \frac{π}{6} + 2 πn

合并区间

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

合并重叠的区间

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

流行的例子

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

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

sin^{2} (x) < 0

cos (t) < 0

cot (x) < 0

4cos(x/3+pi/4)+sqrt(12)>= 0

4 cos (\frac{x}{3} + \frac{π}{4}) + 12 \geq 0