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

Question

5581 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-29T11:56:09

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

$5 tan 3x -5=0 or 5 tan 3x = 5 or tan 3x =1$ 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 · 2016-10-29T12:09:50

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

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

How do you solve 5tan3x-5=05tan3x-5=0 and find all solutions in the interval [0,2pi)[0,2pi)?