If x sin a +y sin 2a + z sin3a = sin4a,  x sin b +y sin 2b + z sin3b= sin4b,  x sin c+ y sin 2c +z sin 3c= sin4c, then the roots of the equationt3−z2t2−y+24t+z−x8=0,a,b,c≠nπ are

xsin⁡a+ysin⁡2a+zsin⁡3a=sin⁡4a⇒ xsin⁡a+y×2sin⁡acos⁡a+z×sin⁡a3−4sin2⁡a =2×2sin⁡acos⁡acos⁡2a⇒ x+2ycos⁡a+z3+4cos2⁡a−4 =4cos⁡a2cos2⁡a−1 [ as sin⁡a≠0]⇒ 8cos3⁡a−4zcos2⁡a−(2y+4)cos⁡a+(z−x)=0⇒ cos3⁡a−z2cos2⁡a−y+24cos⁡a+z−x8=0Which shows that cos a is root of the equation t3−z2t2−y+24t+z−x8=0Similarly, from second and third equations, we can slow that cos b and cos c are the roots of the given equation.