How can I solve sin(2x) = sin(x/2) for all values of x?

Question

I can find x = 0, and also can show that cos x = (1/4)*(-1+/-sqrt(5)), but I don't know how I should have known that the latter cosines of x correspond to angles involving fifths of pi (radians)? In other words, all I really want to know at this point is if there is a way to recognize the arccos of (1/4)*(-1+/-sqrt(5)).

Answer 1

One way to do this is to think about the unit circle. For the sin values to be equal for both sin (2x) and sin (x/2), the y coordinates of the unit circle need to be the same, or the measure of the angles in radians need to be the same when measured from the 0 angle and the pi angle.  So, you are looking for x, such that x and 4x are angles with the same coordinates.    or you need x + 4x to = pi.  pi/5 and 4* pi/5 are the only such angles, as 5x = pi.  These have the same y coordinates, or sin values.

 

Similarly, the same holds for -pi/5 and -4*pi/5, as x + 4x can also = -pi, asn -pi = pi. 

 

Of course, 0 is an answer as well. 

 

I hope this helps.  The best way to visualize it and reinforce the concept is to draw the unit circle and reference it.