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Popular >

sin(2x)0<= pi/2 \land 0-pi/2 <0

Solution

sin (2 x) 0 \leq \frac{π}{2} and 0 - \frac{π}{2} < 0

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

sin (2 x) \cdot 0 \leq \frac{π}{2} and 0 - \frac{π}{2} < 0

sin (2 x) \cdot 0 \leq \frac{π}{2} :

True for all

x

sin (2 x) \cdot 0 \leq \frac{π}{2}

Apply rule

0 \cdot a = 0

0 \leq \frac{π}{2}

Therefore, the solution is

True for all x

0 - \frac{π}{2} < 0 :

True

0 - \frac{π}{2} < 0

Simplify

0 - \frac{π}{2} : - \frac{π}{2}

0 - \frac{π}{2}

0 - \frac{π}{2} = - \frac{π}{2}

= - \frac{π}{2}

- \frac{π}{2} < 0

Therefore, the solution is

True

Combine the intervals

True for all x \in R and True for all x \in R

Merge Overlapping Intervals

True for all x \in R and True for all x \in R

The intersection of two intervals is the set of numbers which are in both intervals

True for all

x \in R

and

True for all

x \in R

True for all x \in R

True for all x \in R

[sin (π) t] 2 \leq t \leq 4

3 - 3 (0)^{2} \leq g (0) \leq 3 cos (0)

0 < sin (x) cos (x) < 2

csc (θ) < 0 and (csc (θ)) (cot (θ)) > 0

1 > arctan (x) > 0