Question

If ${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x=\frac{3}{8},$ then the value of $8\sin 4x$ is

Answer 1

${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x=\frac{3}{8}$
${\sin }^{3}x(4{\cos }^{3}x-3\cos x)+{\cos }^{3}x(3\sin x-4{\sin }^{3}x)=\frac{3}{8}$
$\Rightarrow 3\sin x\cos x({\cos }^{2}x-{\sin }^{2}x)=\frac{3}{8}$
$\Rightarrow \sin 2x\cos 2x=\frac{1}{4}$
$\Rightarrow \sin 4x=\frac{1}{2}$
$\Rightarrow 8\sin 4x=4$