Question

Find roots of the following cubic equation :
$2x^3-3x^{2}cos(A-B)-2xco{s}^2(A+B)+sin2A.sin2Bcos(A-B)=0.$

Answer 1

$sin2Asin2B=\frac{1}{2}\left[cos\right(2A-2B)-cos(2A-2B\left)\right]$
$=\frac{1}{2}\left[2co{s}^2\right(A-B)-1-2co{s}^2(A+B)+1]$
$2x^3-3x^{2}cos(A-B)-2xco{s}^2(A+B)+co{s}^3(A-B)-co{s}^2(A+B)cos(A-B)=0$
We write the above equation as under :
$2x^3+x^{2}cos(A-B)-4x^{2}cos(A-B)+co{s}^3(A-B)-co{s}^2(A+B)2x+cos(A-B)=0$
or $x^{2}2x+cos(A-B)-cos(A-B)4x^{2}co{s}^2(A-B)-co{s}^2(A-B)2x+cos(A-B)=0$
Clearly $2x+cos(A-B)$ is a factor
$\therefore 2x+cos(A-B)[x^2+cos(A-B\left)\right]2x-cos(A-B)-co{s}^2(A-B)=0$
or $\therefore 2x+cos(A-B)[x^2+cos(A-B\left)\right]+co{s}^2(A-B)-co{s}^2(A+B)=0$
The first factor gives $x=-\frac{1}{2}cos(A-B)=0$
Second factor by solving as a quadratic gives
$x=\frac{2cos(A-B)\pm 2cos(A+B)}{2}$
$=2cosAcosBor2sinAsinB$
Hence the roots are
$2cosAcosB,2sinAsinBand-\frac{1}{2}cos(A-B)$