Animated Solution for Mathematics - Trigonometry: The sum of the solutions $x\in \mathbb{R}$ of the equation $\frac{3\cos 2x+{\cos }^{3}2x}{{\cos }^{6}x-{\sin }^{6}x}=x^3-x^2+6$ is

Imagine you are standing before a massive, intimidating trigonometric expression. The equation

seems designed to make you panic. In the world of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back that mask together.

Our first mission is to simplify the left-hand side. The numerator, $3\cos 2x+{\cos }^{3}2x$, can be factored as:

Now, look at the denominator: ${\cos }^{6}x-{\sin }^{6}x$. This is a classic difference of cubes, $({\cos }^{2}x{)}^3-({\sin }^{2}x{)}^3$.

Using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, we expand this into:

The first bracket is the identity $\cos 2x$. The second bracket, ${\cos }^{4}x+{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x$, can be rewritten by completing the square:

Using the double angle identity $\sin 2x=2\sin x\cos x$, we know ${\sin }^{2}x{\cos }^{2}x=\frac{1}{4}{\sin }^{2}2x$. Thus, the denominator becomes:

Now, bring the numerator and denominator together:

The $\cos 2x$ terms cancel out. We are left with:

If we convert ${\sin }^{2}2x$ to $1-{\cos }^{2}2x$, the denominator becomes $\frac{3}{4}+\frac{1}{4}{\cos }^{2}2x$, which is $\frac{1}{4}(3+{\cos }^{2}2x)$. The entire expression collapses to 4.

We are left with the simple cubic equation $4=x^3-x^2+6$, or:

By testing small integers, we find that $x=-1$ is a root: $(-1{)}^3-(-1{)}^2+2=-1-1+2=0$. Dividing the cubic by $(x+1)$ gives us the quadratic factor $x^2-2x+2$.

The discriminant of this quadratic is $D=(-2{)}^2-4(1)(2)=-4$. Since $D<0$, there are no further real roots.

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