tan(x)<= sqrt(3)

解答

tan (x) \leq 3

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

间隔符号

(- \frac{π}{2} + πn, \frac{π}{3} + πn]

十进制

- 1.57079 \dots + πn < x \leq 1.04719 \dots + πn

求解步骤

tan (x) \leq 3

若

tan (x) \leq a

，则

- \frac{π}{2} + πn < x \leq arctan (a) + πn

- \frac{π}{2} + πn < x \leq arctan (3) + πn

化简

arctan (3) : \frac{π}{3}

arctan (3)

使用以下普通恒等式:

arctan (3) = \frac{π}{3}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{3}

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

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

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

cos (2 x) \geq 1

cos (x) - sin (x) \geq - 1

sin (x) - 3 cos (x) > 0