Question

How do you solve `5tan3x-5=0` and find all solutions in the interval `[0,2pi)`?

Answer 1

In the given interval Solution: `x =pi/12; x=(5pi)/12; x= (9pi)/12 ; x= (13pi)/12 ; x= (17pi)/12 ; x= (21pi)/12 ;`

Explanation:

`5 tan 3x -5=0 or 5 tan 3x = 5 or tan 3x =1` We know `tan (pi/4)=1 :.3x =pi/4` also `tan (pi+pi/4)=1 or tan ((5pi)/4)=1 :. 3x=(5pi)/4`

The given interval for `x` is `[0,2pi)`. New interval for `3x` is `[0,6pi)`. so in the interval `[0,6pi)`.

`3x =pi/4; 3x=(5pi)/4; 3x= pi/4+2pi=(9pi)/4 ; 3x= (5pi)/4+2pi=(13pi)/4 ; 3x= pi/4+4pi=(17pi)/4 ; 3x= (5pi)/4+4pi=(21pi)/4 ;`

Solution:`x =pi/12; x=(5pi)/12; x= (9pi)/12 ; x= (13pi)/12 ; x= (17pi)/12 ; x= (21pi)/12 ;` [Ans]

Answer 2

As `tan 3x =1 = tan( pi/4)`, the general solution is

`3x=kpi+pi/4 to x = kpi/3+pi/12, k = 0, +-1, +-2, +-3, ...`

Upon setting, k = 0, 1, 2, 3, 4 and 5

`x = pi/4, 5/4pi, 9/4pi, 13/4pi,17/4pi and 21/4pi in [0, 2pi]`.