JEE 2023 — Mathematics PYQ

$\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-\sin 6x}=x^3-x^2+6$

L.H.S

$\frac{\cos 2x(3+{\cos }^{2}2x)}{\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{3}x+{\sin }^{4}x+{\cos }^{2}x{\sin }^{2}x\right)}$

$\therefore \cos 2x={\cos }^{2}x-{\sin }^{2}2x$

$\frac{\cos 2x(3+{\cos }^{2}2x)}{\cos 2x\left({\left({\cos }^{2}x+{\sin }^{2}x\right)}^2-{\cos }^{2}x{\sin }^{2}x\right)}$

$\frac{3+{\cos }^{2}24}{1-\frac{1}{4}{\sin }^{2}2x}$

$\therefore \sin 2x=2\sin x\cos x$

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

$=4\frac{4-{\sin }^{2}2x}{4-{\sin }^{2}2x}$

Now $\frac{x^3-x^2+6}{x^3-x^2+2}=0$

∴ sum of solution = 1