Pan African MO

To solve the given equation for $x \in \mathbb{R}$:
$$\sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x},$$
we start by simplifying the expression.

### Step 1: Simplify the given equation
We know from the trigonometric identities:
$$\cot{x} = \frac{\cos{x}}{\sin{x}} \quad \text{and} \quad \tan{x} = \frac{\sin{x}}{\cos{x}}.$$
Substituting these into the equation, we have:
$$\sin^3{x}\left(1 + \frac{\cos{x}}{\sin{x}}\right) + \cos^3{x}\left(1 + \frac{\sin{x}}{\cos{x}}\right) = \cos{2x}.$$

Simplifying:
$$\sin^3{x} + \cos{x}\sin^2{x} + \cos^3{x} + \sin{x}\cos^2{x} = \cos{2x}$$

Combining terms:
$$\sin^3{x} + \cos^3{x} + \sin{x}\cos^2{x} + \cos{x}\sin^2{x} = \cos{2x}$$

### Step 2: Use Trigonometric Identities
Use the identity for sum of cubes, $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$:
$$a = \sin{x}, \quad b = \cos{x}$$

Since $\sin^2{x} + \cos^2{x} = 1$, the term $(\sin{x} + \cos{x})(1 - \sin{x}\cos{x})$ simplifies part of our identity:
$$\sin^3{x} + \cos^3{x} = (\sin{x} + \cos{x})(\sin^2{x} - \sin{x}\cos{x} + \cos^2{x})$$

Simplifies further to:
$$\sin^3{x} + \cos^3{x} = (\sin{x} + \cos{x})(1 - \sin{x}\cos{x})$$
Thus:
$$(\sin{x} + \cos{x})(1 - \sin{x}\cos{x}) + \sin{x}\cos{x}(\sin{x} + \cos{x}) = \cos{2x}$$

Factor out $(\sin{x} + \cos{x})$:
$$(\sin{x} + \cos{x})(1 - \sin{x}\cos{x} + \sin{x}\cos{x}) = \cos{2x}$$

Simplify to:
$$(\sin{x} + \cos{x}) = \cos{2x}$$

### Step 3: Solve for $x$
Using the identity for the double angle:
$$\cos{2x} = \cos^2{x} - \sin^2{x} = 2\cos^2{x} - 1$$

Equating:
$$\sin{x} + \cos{x} = 2\cos^2{x} - 1$$

Let $u = \sin{x} + \cos{x}$, and given that $(\sin{x} + \cos{x})^2 = \sin^2{x} + 2\sin{x}\cos{x} + \cos^2{x} = 1 + 2\sin{x}\cos{x}$,
$$u^2 = 1 + 2\sin{x}\cos{x}$$

For solutions:
$$\sin{x} + \cos{x} = 0 \Rightarrow \sin{x} = -\cos{x} \Rightarrow \tan{x} = -1$$

The solutions occur at:
$$x = m\pi - \frac{\pi}{4}, \quad m \in \mathbb{Z}.$$

Thus, the solution for $x$ is:
$$\boxed{x = m\pi - \frac{\pi}{4}, \ m \in \mathbb{Z}}$$