Question 1199347
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If tan(x/2)=sin(x) and cos(x)≠0, then the value of tan(x) is . . . 
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<pre>
Our starting equation is 

    {{{tan(x/2)}}} = {{{sin(x)}}}.      (1)


Since {{{tan(x/2)}}} is not defined when {{{cos(x/2)}}} = 0, the domain of this equation is 
the set of all real numbers except {{{x/2}}} = {{{pi/2+k*pi}}},  or  x = {{{pi+2k*pi}}} = {{{(2k+1)*pi}}}.


One obvious solution to equation (1) is x = {{{2k*pi}}}, when both sides of equation (1) are zeroes.    (2)


In what follows, we assume that x is neither {{{2k*pi}}} nor {{{(2k+1)*pi}}}.    (3)


        In particular, it implies that {{{sin(x/2)}}} =/= 0.    (4)


Now we transform equation (1) step by step

    {{{sin(x/2)/cos(x/2)}}} = {{{sin(2*(x/2))}}}

    {{{sin(x/2)/cos(x/2)}}} = {{{2sin(x/2)cos(x/2)}}}


Due to (4), we can divide both sides by {{{sin(x/2)}}}, since this factor is not zero.

    {{{1/cos(x/2)}}} = {{{2cos(x/2)}}}

    {{{1}}} = {{{2cos^2(x/2)}}}

    {{{1/2}}} = {{{cos^2(x/2)}}}

    {{{cos(x/2)}}} = {{{""+-sqrt(1/2)}}} = {{{""+-sqrt(2)/2}}}

    {{{x/2}}} = {{{""+-pi/4 + pi*n}}}    <<<---=== notice here the difference with the Edwin's solution
                                 (here Edwin lost some/(half of the) solutions)


    {{{x}}} = {{{""+-pi/2 + 2pi*n}}} = {{{2pi*n +- pi/2}}} = {{{pi*(2n +- 1/2 )}}}, where n is any integer.

              But for all these values of x, cos(x) = 0, so we should exclude this set of values
              from our consideration due to condition cos(x) =/= 0, imposed in the problem.


Finally, the only set of solutions is  x = {{{2k*pi}}}, found in (2).


<U>ANSWER</U>.  The solution to equation (1) with the imposed condition cos(x) =/= 0 

         is the set  x = {{{2k*pi}}}, and tan(x) = 0 for all these values of x.
</pre>

Solved.