Question 1075988
.
Can you please help me solve this equation: {{{ sin(x+1)=cos(x) }}} with answers between 0 and 2pi?
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<pre>
sin(x+1) = cos(x)  <---->  {{{sin(x+1)}}} = {{{sin(pi/2 - x)}}}  


This equation has TWO roots in the area of interest:

1)  x+1 = {{{pi/2-x}}}  ---->  2x = {{{pi/2 -1}}}  <---->  x = {{{(1/2)*(pi/2-1)}}},

     and 

2)  x+1 = {{{pi/2-x + 2*pi}}}  ---->  2x = {{{pi/2 -1 + 2pi}}}  <---->  x = {{{(1/2)*(pi/2-1)}}} + {{{pi)}}}.
</pre>

<U>Answer</U>.  The solutions are  x = {{{(1/2)*(pi/2-1)}}} and x = {{{(1/2)*(pi/2-1)}}} + {{{pi)}}}.



{{{graph( 330, 330, -1.5, 6.5, -1.5, 1.5,
          sin(x+1), cos(x)
)}}}


Plots y = sin(x+1) (red) and y = cos(x) (green)



The plot shows two roots in the interval [0, 2pi).


Solved.