Question 1103336
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<pre>
sin(2x) = 1  ====>

2x = {{{pi/2}}}.


The last equation 2x = {{{pi/2}}} has TWO solutions in the interval [0,{{{2pi}}}).


One of these two solutions is  {{{x[1]}}} = {{{pi/4}}},  which is OBVIOUS.


The second solution  is  <U>to add</U>  {{{pi}}}  to the first solution:

    {{{x[2]}}} = {{{pi/4 + pi}}} = {{{5pi/4}}}.


Why ?  - Because when you take  {{{2*x[2]}}}  with  {{{x[2]}}} = {{{pi/4 + pi}}},

you will get  {{{2*x[2]}}} = {{{pi/2 + 2pi}}},


which is <U>GEOMETRICALLY THE SAME</U> AS the ANGLE {{{pi/2}}}.


So your  {{{2*x[2]}}}  is geometrically the same  {{{pi/2}}}  and satisfies the original equation  sin(2x) = 1.


Notice that both  {{{x[1]}}} = {{{pi/4}}}  and  {{{x[2]}}} = {{{pi/4 + pi}}} belong to the interval [0,{{{2pi}}}), so they both are the solutions 

to your original equation  sin(2x) = 1  in this interval.



The plot below confirms my arguments:


{{{graph( 330, 330, -0.5, 6.5, -1.5, 1.5,
          sin(2x), 1
)}}}


Plot y = sin(2x)  (red)  and  y = 1 (green)
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