Posted by wendell on Friday, June 8, 2012 at 2:24pm.
Assuming there is no typo in the question, we have to solve:
4=2cos(pi/6t-2pi/3)+4
the 4 on each side cancel out, so the equation becomes:
2cos(pi/6t-2pi/3) = 0
Divide by 2 on each side to get:
cos(pi/6t-2pi/3) = 0
cos(x)=0 when x=(k+1/2)π, k∈Z
(i.e. k is an integer).
Therefore we get:
(π/6)t-2π/3=(k+1/2)π
Divide by π on both sides and solve for t:
t/6-2/3=(k+1)/2
t/6 = k/2 + 1/2 + 2/3
t/6 = k/2 + 7/6
Multiply both sides by 6:
t = 3k + 7 where k∈Z
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