xsin⁡a+ysin⁡2a+zsin⁡3a=sin⁡4ax \sin a + y \sin 2a + z \sin 3a = \sin 4a xsin⁡b+ysin⁡2b+zsin⁡3b=sin⁡4bx \sin b + y \sin 2b + z \sin 3b = \sin 4b…

Problem

xsin⁡a+ysin⁡2a+zsin⁡3a=sin⁡4ax \sin a + y \sin 2a + z \sin 3a = \sin 4a xsin⁡b+ysin⁡2b+zsin⁡3b=sin⁡4bx \sin b + y \sin 2b + z \sin 3b = \sin 4b xsin⁡c+ysin⁡2c+zsin⁡3c=sin⁡4cx \sin c + y \sin 2c + z \sin 3c = \sin 4c . Then, the roots of the equation. t3−z2t2−y+24t+z−x8=0t^3 - \frac{z}{2}t^2 - \frac{y+2}{4}t + \frac{z-x}{8} = 0 ( a,b,c≠nπa, b, c \neq n\pi ) are a. sin⁡a,sin⁡b,sin⁡c\sin a, \sin b, \sin c b. cos⁡a,cos⁡b,cos⁡c\cos a, \cos b, \cos c c. sin⁡2a,sin⁡2b,sin⁡2c\sin 2a, \sin 2b, \sin 2c d. cos⁡2a,cos⁡2b,cos⁡2c\cos 2a, \cos 2b, \cos 2c

Browse the problem library