Question 1069081
.
Give the general solution to the equation sinx+sin2x+sin3x=0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
sin(x) + sin(2x) + sin(3x) = 0.     (1)


Apply the trigonometry formula  {{{sin(a) + sin(b)}}} = {{{2*sin((a+b)/2)*cos((a-b)/2)}}}

     (see any serious textbook in trigonometry or the lessons 

         - <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Compendium-of-Trigonometry-Formulas.lesson>FORMULAS FOR TRIGONOMETRIC FUNCTIONS</A>
         - <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-of-trigonometric-functions.lesson>Addition and subtraction of trigonometric functions</A>

      in this site) to the first and third addend in the left side of the original equation (1).   You will get

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

Now, the equation (1) takes the form

{{{2*sin(2x)*cos(x) + sin(2x)}}} = {{{0}}},   or

{{{sin(2x)*(2*cos(x) +1)}}} = {{{0}}}.

This equation deploys in two independent equations


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


2.  2cos(x) + 1 = 0  --->  cos(x) = {{{-1/2}}}  --->  x = {{{2pi/3 + 2k*pi}}},  x = {{{4pi/3 + 2k*pi}}},  k = 0, =/-1, +/-2. . . . 


<U>Answer</U>.  The solutions are  a)  x = {{{k*pi}}}, x = {{{pi/2 + k*pi}}},  k = 0, =/-1, +/-2. . . .    and 

                            b)  x = {{{2pi/3 + 2k*pi}}},  x = {{{4pi/3 + 2k*pi}}},  k = 0, =/-1, +/-2. . . . 


{{{graph( 330, 330, -1.5, 6.5, -2.5, 2.5,
          sin(x)+sin(2*x)+sin(3*x)
)}}}


Plot y = sin(x) + sin(2x) + sin(3x)
</pre>

The solution by the other tutor is incorrect.