Question 1138303
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sin(x) + sin(2x) + sin(3x) + sin(4x) = 0.        (1)


Use the general formula of Trigonometry

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


You have 

sin(x) + sin(4x) = {{{2*sin((x+4x)/2)*cos((4x-x)/2)}}} = {{{2*sin(2.5x)*cos(1.5x)}}},

sin(2x) + sin(3x) = {{{2*sin((2x+3x)/2)*cos((3x-2x)/2)}}} = {{{2*sin(2.5x)*cos(0.5x)}}}.


Therefore, the left side of the original equation is

sin(x) + sin(2x) + sin(3x) + sin(4x) = 2*sin(2.5x)*cos(1.5x) + 2*sin(2.5x)*cos(0.5x) = 2*sin(2.5x)*(cos(1.5x) + cos(0.5x)).


Hence, the original equation is equivalent to

2*sin(2.5x)*(cos(1.5x) + cos(0.5x)) = 0,    or,  canceling  the factor  2*sin(2.5x),

cos(1.5x) + cos(0.5x) = 0.                       (3)


Next, apply another general formula of Trigonometry

{{{cos(a) + cos(b)}}} = {{{2cos((a+b)/2)*cos((a-b)/2)}}}.                (4)


Then the equation (3) becomes

{{{2*cos(x)*cos(x/2)}}} = 0.                                (5)


Equation (5) deploys in two independent separate equations:


1.  cos(x) = 0  --->  x = {{{pi/2 + k*pi}}},  k = 0, +/-1, +/-2, . . . 


2.  cos(x/2) = 0  --->  {{{x/2}}} = {{{pi/2 + k*pi}}},  k = 0, +/-1, +/-2, . . . ,  or

                        x = {{{pi + 2k*pi}}} = {{{(2k+1)*pi}}},  k = 0, +/-1, +/-2, . . . 


From (1) and (2), in the given interval the original equation has the roots {{{pi/2}}}, {{{pi}}}, {{{3pi/2}}},  or  90°,  180°,  270°.


<U>But these are not ALL the roots</U>.

There is one more family of roots.

Do you remember I canceled the factor 2*sin(2.5x) ?

Of course, I must consider (and add to the solution set !) all the solutions of the equation

sin(2.5x) = 0.

They are  2.5x = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 

or, which is the same,

{{{(5x)/2}}} = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = 0, {{{(2k*pi)/5}}}, {{{(4k*pi)/5}}}, {{{(6k*pi)/5}}}, {{{(8k*pi)/5}}},  k = 0, +/-1, +/-2, . . . 

<U>The final answer is</U>:  There are two families of solutions. 

                      One family is  {{{pi/2 + 2k*pi}}},  {{{pi+2k*pi}}},  and  {{{3pi/2+ 2k*pi}}},  k = 0, +/-1, +/-2, . . . , or  90°, 180° and 270°.

                      The other family is  {{{(2k*pi)/5}}},  k = 0, +/-1, +/-2, . . . ,  or  0°, 72°, 144°, 216°, 288°.
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Solved.


<U>CHECK</U>


See the plot of the left side of the original equation


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{{{graph( 330, 330, -1, 6.5, -5, 5,
          sin(x) + sin(2x) + sin(3x) + sin(4x)
)}}}


Plot y = sin(x) + sin(2x) + sin(3x) + sin(4x)



Do you see 8 roots in the interval [{{{0}}},{{{2pi}}}) ?
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