Question 1009490
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Solve {{{tan(theta/2)}}} - {{{sin(theta)}}} = {{{0}}}.
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The half-argument formula says that {{{tan(theta/2)}}} = {{{sin(theta)/(1+cos(theta))}}}. 

(See, for example, the lesson <A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A> in this site).


Substitute it into the original equation, and you will get

{{{sin(theta)/(1+cos(theta))}}} - {{{sin(theta)}}} = {{{0}}}.

Now factor it:

{{{sin(theta)}}}.{{{(1/(1+cos(theta))-1)}}} = {{{0}}}.

Thus you get two equations:

1) {{{sin(theta)}}} = {{{0}}},  which has two solutions  {{{theta}}} = {{{0}}} and {{{pi}}} in the given interval for {{{theta}}}.


2) {{{1/(1+cos(theta))-1}}} = {{{0}}},  or  {{{cos(theta)}}} = 0,  which has two solutions  {{{theta}}} = {{{pi/2}}} and {{{3pi/2}}} in the given interval for {{{theta}}}.

<U>Answer</U>. The solutions are {{{theta}}} = {{{0}}}, {{{pi/2}}}, {{{pi}}} and {{{3pi/2}}}.
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