A block is attached to a spring and set in motion on a frictionless plane. Its location on the surface at any time "t" in seconds is given in meters by x= sqrt3 sin 2t + cos 2t. For what values of t is the block at its resting position? x=0
+33616 We want the times for which (√3)sin(2t) + cos(2t) = 0. We can rearrange this to get tan(2t) = -1/√3 (by dividing through by √3*cos(2t) and moving one term to the other side of the = sign).
Now tan(pi/6) is 1/√3, but it is negative in the second and fourth quadrants, so we must have tan(n*pi - pi/6) = -1/√3, where n is a positive integer and the angles are in radians (in degrees pi is 180 and pi/6 is 30).
This means 2t = n*pi - pi/6 or t = n*pi/2 - pi/12.
+33616 We want the times for which (√3)sin(2t) + cos(2t) = 0. We can rearrange this to get tan(2t) = -1/√3 (by dividing through by √3*cos(2t) and moving one term to the other side of the = sign).
Now tan(pi/6) is 1/√3, but it is negative in the second and fourth quadrants, so we must have tan(n*pi - pi/6) = -1/√3, where n is a positive integer and the angles are in radians (in degrees pi is 180 and pi/6 is 30).
This means 2t = n*pi - pi/6 or t = n*pi/2 - pi/12.
