JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Trigonometric Ratios and Identities

${\cos }^{6}x-{\sin }^{6}x=({\cos }^{2}x-{\sin }^{2}x)({\cos }^{4}x+{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x)=\cos 2x\left(1-{\cos }^{2}x{\sin }^{2}x\right)$

$=\cos 2x\left(1-\frac{{\sin }^{2}2x}{4}\right)=\cos 2x\cdot \frac{4-{\sin }^{2}2x}{4}=\cos 2x\cdot \frac{4-(1-{\cos }^{2}2x)}{4}=\frac{\cos 2x(3+{\cos }^{2}2x)}{4}$.

The numerator is $3\cos 2x+{\cos }^{3}2x=\cos 2x(3+{\cos }^{2}2x)$, exactly the bracket above times $\cos 2x$, so

$\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=\frac{\cos 2x(3+{\cos }^{2}2x)}{\frac{\cos 2x(3+{\cos }^{2}2x)}{4}}=4$ (wherever $\cos 2x(3+{\cos }^{2}2x)\ne 0$).

So the equation becomes $4=x^3-x^2+6\Rightarrow x^3-x^2+2=0$.

$x=-1$ is a root: $(-1{)}^3-(-1{)}^2+2=-1-1+2=0$.

Factor: $x^3-x^2+2=(x+1)(x^2-2x+2)$. The quadratic $x^2-2x+2=0$ has discriminant $4-8=-4<0$, so no real roots there.

The only real solution is $x=-1$, so the sum of real solutions is $-1$.