So,y(t) = 2.5e^-t cos2t

Question

I need to find the derivative, which is
y'(t)= -2.5e^-t(2sin 2t +cost 2t) .
And now I need to find t when y'(t) = 0
(I know 2.5 e^-t is never zero.)
I need to use trigonometry identity to find it ?

Answer 1

so you are solving

2sin 2t + cos 2t = 0
2sin 2t = -cos 2t
2sin 2t/cos 2t = -1
tan 2t = -1/2

set your calculator to RAD
and find 2nd function tan
(tan^-1 (+1/2)
to get .46365
we know that the tangent is negative in II and IV
so 2t = π - .46365 = 2.6779
t = 1.339

or

2t = 2π - .46365 = 5.8195
t = 2.9098

We also know that the period of tan 2t = 2π/2 = π
so by adding or subtracting multiples of π to any existing answer will yield as many answers as you want.
e.g
t = 1.339 + π = appr 4.4806
t = 2.9098 + 5π = 18.6177

checking with that last value
2sin(2(18.6177)) + cos(2(18.6177))
= very very close to zero

Answer 2

ok~ get it clearly..thanks^^

Answer 3

just recall that the period of tan(t) is pi, not 2pi, so

tan(2t) has period pi/2