Expert: Ahmed Salami - 12/17/2005

Question
For 0<è<2ð,
how do I go about solving all possible solutions for:
tan3è(tan3è + 1) = 2?
Thank you very much in advance (I could do all other questions on trig homework, but don't know how to involve tan)
Daniel

Answer
Hi Daniel,
Sorry for the time taken.
For tan3e(tan3e + 1) = 2
let tan3e = x and we have
x(x+1) = 2
x^2 + x - 2 = 0
(x+2)(x-1) = 0
x = -2 or 1
Therefore
tan3e = -2 or 1
For tan3e = -2
3e = -63.44
e = -21.15
But all the solutions for the trigonometric equation
tan# = y are
# = P + n(pi)
where p is the first value from calculation and n is an
integer positive, negative or zero. Also, we are working
in degrees so pi = 180
So our sequence of solutions is
e = ....., -21.15, -21.15 + pi, -21.15 + 2pi, ....
but because of the limitation 0<e<2pi, we can only take values
-21.15 + 180 and -21.15 + 360
i.e 158.85, 338.85
For the second solution tan3e = 1
3e = 45
e = 15
Proceeding as we did earlier, the sequence is
....., 15, 15 + pi, 15 + 2pi, ...
e = 15 and 15 + 180
i.e 15, 195
Therefore, for the restriction 0<e<2pi
The complete solution of tan3e(tan3e + 1) = 2 are
e = 15, 158.85, 195 and 338.85
I hope i have helped, you can always get back to me.
Regards.