SOLUTION: Identify the solution(s) of tan(x+π)−tan(π−x)=0 for 0≤x<2π.

Algebra ->  Trigonometry-basics -> SOLUTION: Identify the solution(s) of tan(x+π)−tan(π−x)=0 for 0≤x<2π.      Log On


   

Question 1179360: Identify the solution(s) of tan(x+π)−tan(π−x)=0 for 0≤x<2π.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Identify the solution(s) of
tan%28x%2Bpi%29-tan%28pi-x%29=0 for+0%3C=x%3C2pi
rewrite using trig identities
tan%28x%2Bpi%29=sin%28x%2Bpi%29%2Fcos%28x%2Bpi%29
tan%28pi-x%29=sin%28pi-x%29%2Fcos%28pi-x%29

sin%28x%2Bpi%29%2Fcos%28x%2Bpi%29-sin%28pi-x%29%2Fcos%28pi-x%29=0........using trig identities

sin%28x%2Bpi%29=sin%28x%29cos%28pi%29%2Bcos%28x%29sin%28pi%29
sin%28pi-x%29=sin%28pi%29cos%28x%29-cos%28pi%29sin%28x%29
cos%28x%2Bpi%29=cos%28x%29cos%28pi%29-sin%28x%29sin%28pi%29
cos%28pi-x%29=cos%28pi%29cos%28x%29%2Bsin%28pi%29sin%28x%29



substitute: sin%28pi%29=0, cos%28pi%29+=-1




%28-sin%28x%29%29%2F%28-cos%28x%29%29%0D%0A-%28sin%28x%29%29%2F%28-cos%28x%29%29

sin%28x%29%2Fcos%28x%29%0D%0A%2Bsin%28x%29%2Fcos%28x%29

2tan%28x%29+=0

x=tan%5E-1%280%29
x=0°
tan%28x%29 is periodic, the period is pi
then next solution is x=0%2B180=180°
so, in interval +0%3C=x%3C2pi solutions are
x=0°
x=180°


Answer by ikleyn(52908) About Me  (Show Source):
You can put this solution on YOUR website!
.
Identify the solution(s) of tan(x+π)−tan(π−x)=0 for 0≤x<2π.
~~~~~~~~~~~~~~ 

Your starting equation is 

    tan(x+π)−tan(π−x) = 0    for 0≤x<2π.


It is the same as  (is equivalent to)

    tan(x+π) = tan(π−x)      for 0≤x<2π.


Since tan is a monotonic periodic function with the period of  π, it means that

    (x+π) - (π−x) = kπ,   where  k = 0, +/-1, +/-2, . . . 

or

    2x = kπ,    where  k = 0, +/-1, +/-2, . . . 



Having x in the interval  0≤x<2π,  it means that  

    x = 0  or  x = π.        ANSWER

Solved.