The number of real roots of the equation 2cos(x(x + 1)) = 2x + 2-x is

For any positive real number y, the AM–GM inequality gives y + 1/y ≥ 2, with equality exactly when y = 1. Separately, since the cosine function is bounded above by 1, any expression of the form 2cos(θ) never exceeds 2. These two one-sided bounds — one always ≥ 2, the other always ≤ 2 — can only meet where both equal exactly 2.

1. Let y = 2x (y > 0, since an exponential is always positive). The right-hand side becomes 2x + 2-x = y + 1/y, so by AM–GM, RHS ≥ 2, with equality only when y = 1/y, i.e. y = 1, i.e. x = 0.

2. The left-hand side is 2cos(x(x + 1)). Since cos(θ) ≤ 1 for every real θ, LHS ≤ 2 for every x.

3. For LHS = RHS to hold, a value that is simultaneously ≥ 2 (from RHS) and ≤ 2 (from LHS) is needed — the only possibility is LHS = RHS = 2.

4. RHS = 2 forces x = 0 (from Step 1, the unique equality case).

Substituting x = 0 back into both sides confirms this: LHS = 2cos(0 · 1) = 2cos(0) = 2, and RHS = 20 + 20 = 1 + 1 = 2 — both sides equal 2, so x = 0 does satisfy the equation.

Hence the equation has exactly one real root.