Question 1210192
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Find an acute angle A such that sin 4A = sin A + sin 2A. Express your answer in degrees.
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        In this my post,  I will give a standard solution to this problem.


        Saying " a standard solution ",  I mean a tradition method,  by which I 

        (and thousands and millions other students of my generation)  solved such problems in a class, 
        considering it as a routine.



<pre>
Start from

    sin(4A) = sin(A) + sin(2A).    (1)


Transform into

    sin(4A) - sin(2A) = sin(A).


Use the common/general formula of Trigonometry

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

valid for any angles 'a' and 'b'.  Apply it for a = 4A, b = 2A.  
You will get

    2cos(3A)*sin(A) = sin(A).    (2)


One solution is sin(A) = 0,  which implies  A = 0 or {{{pi}}}, 
which in degrees is A = 0° or A = 180°.


Next, if we want to find other solutions with sin(A) =/= 0, then 
we can cancel sin(A) in both sides of equation (2).  You will get then

    2cos(3A) = 1,  or  cos(3A) = {{{1/2}}}.


It has the solutions

    3A = {{{pi/3 + 2k*pi}}}  or  3A = {{{5pi/3+2k*pi}}},  k = 0. +/-1, +/-2, . . . 

which implies

    A = {{{pi/9+(2/3)*k*pi}}}  or  A = {{{5pi/9+(2/3)*k*pi}}},  k = 0. +/-1, +/-2, . . . 
  


In degrees, it is

    A = 20° + 120°*k  or  A = 100° + 120°*k.


<U>ANSWER</U>.  The solutions are  k*180°,  20° + 120*k,  100° + 120*k,  k = 0, +/-1, +/-2, . . . 


         If you want to list the angles in the interval [0°,360°), they are

         0°, 20°, 100°, 140°, 220°, 260°, 340°.


         An acute angles-solutions are 0° (degenerated) and 20°.
</pre>

Solved.



Thus I got all the solutions, without missing no one.



I prepared a plot of both sides of the original equation (1) in the interval from  {{{-2pi}}}  to  {{{2pi}}}.  
See the link


<A HREF=https://www.desmos.com/calculator/cgohxl603j>https://www.desmos.com/calculator/cgohxl603j</A> 


https://www.desmos.com/calculator/cgohxl603j



Looking at this plot, you may see all the roots as the intersections of the plotted curves.

You may count the number of the roots in the interval [0°,360°), 
which will give you an additional visual check to my solution.