How to solve sin (3x) = sin (x) in the interval [0,2pi)

Replace in the right side of the equation sin2 x by (1 - cos2 x), then solve the quadratic equation for cos x.

cosx - 2=1 - cos2 x

cos2 x + cosx - 3=0

Use the new improved quadratic formula (Socratic Search)

D = d2 = b2 - 4ac=1+12=13 - - > d=±13

There are 2 real roots:

cosx = - b 2a ± d 2a = - 12 ± √13 2 =

Computing by calculator:

cos x = - 0.5 + 1.80 = 1.30

cos x = - 0.5 - 1.80 = - 2.30

Both solutions are rejected (because - 1 < x < 1). The equation is not true.