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[sin(pi)t]2<= t<= 4

Solution

[sin (π) t] 2 \leq t \leq 4

0 \leq t \leq 4

Interval Notation

[0, 4]

Solution steps

[sin (π) t] \cdot 2 \leq t \leq 4

a \leq u \leq b

then

a \leq u and u \leq b

[sin (π) t] \cdot 2 \leq t and t \leq 4

[sin (π) t] \cdot 2 \leq t : t \geq 0

[sin (π) t] \cdot 2 \leq t

Rewrite in standard form

[sin (π) t] \cdot 2 \leq t

Expand

[sin (π) t] \cdot 2 : 0

[sin (π) t] \cdot 2

Remove parentheses:

(a) = a

= sin (π) t \cdot 2

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0 \cdot 2 t

Apply rule

0 \cdot a = 0

= 0

0 \leq t

Subtract

t

from both sides

0 - t \leq t - t

Simplify

- t \leq 0

- t \leq 0

Multiply both sides by

- 1

- t \leq 0

Multiply both sides by -1 (reverse the inequality)

(- t) (- 1) \geq 0 \cdot (- 1)

Simplify

t \geq 0

t \geq 0

Combine the intervals

t \geq 0 and t \leq 4

Merge Overlapping Intervals

t \geq 0 and t \leq 4

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

t \geq 0

and

t \leq 4

0 \leq t \leq 4

0 \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

cosh (θ) = \frac{8}{3} and θ < 0, sinh (θ)