Question 1127695
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What are the solutions for  sin(4x) + sin(x) = 0,  in the interval  [{{{0}}},{{{2pi}}}).
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Please pay attention on how I edited your condition.</U>



<pre>
sin(4x) + sin(x) = 0.      (1)



Use the formula (one of the basic Trigonometry formula)

   sin(a) + sin(b) = {{{2*sin((a+b)/2)*sin((a-b)/2)}}},

which is valid for any angles "a" and "b".



By applying it, you get from (1)


    {{{2*sin(2.5x)*sin(1.5x)}}} = 0.


Thus you must consider two separate cases.



a)  sin(2.5x) = 0  ====>  2.5x = 0,  {{{pi}}},  {{{2pi}}},  {{{3pi}}},  {{{4pi}}},  {{{5pi}}}  ====>  x = 0,  {{{pi/2.5}}},  {{{2pi/2.5}}},  {{{3pi/2.5}}},  {{{4pi/2.5}}},  {{{5pi/2.5}}} = {{{2pi}}} = same as 0,

                          and after that the roots repeat cyclically.

                          This row of solutions is the same as  0,  {{{2pi/5}}},  {{{4pi/5}}},  {{{6pi/5}}},  {{{8pi/5}}}  and {{{10pi/5}}} = {{{2pi}}} = same as 0.


                          In more compact form, this family of solutions is  {{{k*(2pi/5)}}},  k = 0, 1, 2, 3, 4.



b)  sin(1.5x) = 0  ====>  1.5x = 0,  {{{pi}}},  {{{2pi}}},  {{{3pi}}}  ====>  x = 0,  {{{pi/1.5}}},  {{{2pi/1.5}}},  {{{3pi/1.5}}} = {{{2pi}}} = same as 0,

                          and after that the roots repeat cyclically.

                          This row of solutions is the same as  0,  {{{2pi/3}}},  {{{4pi/3}}},  {{{6pi/3}}}= {{{2pi}}} = 0.


                          In more compact form, this family of solutions is  {{{k*(2pi/3)}}},  k = 0, 2, 4.


<U>Answer</U>.  There are two families of solutions:

         a)  {{{k*(2pi/5)}}},  k = 0, 1, 2, 3, 4  and  b)  {{{k*(2pi/3)}}},  k = 0, 2, 4.
</pre>

Solved.