Question 1179360
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Identify the solution(s) of tan(x+π)−tan(π−x)=0 for 0≤x<2π.
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<pre>
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.        <U>ANSWER</U>
</pre>

Solved.