# The velocity function of a moving particle on a coordinate line is v(t) = 3cos(2t) for 0 is less than or equal to t and 2pi is greater than or equal to t (a) Determine when the particle is moving to the right. The particle stops if the velocity is 0, hence, you need to solve for `t in [0,2pi]` the following equation, such that:

`v(t) = 0 => 3cos(2t) = 0 => cos(2t) = 0`

`2t = +-cos^(-1) (0) => 2t = +-pi/2 => t = +-pi/4`

`{(t = pi/4),(t = pi - pi/4 => t = (3pi/4)),(t = pi + pi/4 = (5pi)/4),(t = 2pi - pi/4 = (7pi)/4):}`

You need to evaluate the velocity function over the intervals `[pi/4,(3pi)/4], [(3pi)/4,(5pi)/4],[(5pi)/4,(7pi)/4]` , such that:

`t = pi/2 in [pi/4,(3pi)/4] => 3cos(2(pi/2)) = -3 < 0 =>` the particle moves to the left.

`t = pi in [(3pi)/4,(5pi)/4] => 3cos(2pi) = 3*1 = 3 > 0 =>` the particle moves to the right.

`t = (3pi)/2 in [(5pi)/4,(7pi)/4] => 3cos(2(3pi/2)) = -3 < 0 =>` the particle moves to the left.

Hence, evaluating the moving directions of the particle, under the given conditions, yields that for `t in [(3pi)/4,(5pi)/4],` the particle moves to the right.

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