Question

If $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ and $\alpha \ne n\pi ,n\in \text{I}$ then $864{\sin }^{2}x+3620{\cos }^{2}x$ is equal to

Answer 1

$\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$

$3\sin \alpha -4{\sin }^3\alpha =4\sin \alpha ({\sin }^{2}x-{\sin }^2\alpha )$

$\Rightarrow {\sin }^{2}x=\frac{3}{4}$

$\Rightarrow {\cos }^{2}x=\frac{1}{4}$

$\therefore 864{\sin }^{2}x+3620{\cos }^{2}x=216\times 3+905$

$=648+905$

$=1553$.