Question:

Step 1: Identify the conditions for the equation to be an identity. For a given equation to be an identity (i.e., true for all values of $\theta$, or having more than two roots if it's a polynomial in some variable whose coefficients depend on $\theta$), specific conditions on the coefficients must be met. Based on the problem context, these conditions are given as: $$ \cos 2\theta - 1 = 0 \quad \text{and} \quad 2\cos 2\theta - 1 = 0 $$ These two conditions must be satisfied simultaneously for the equation to be an identity. Step 2: Solve the first condition for $\cos 2\theta$. From the first condition, we isolate $\cos 2\theta$: $$ \cos 2\theta - 1 = 0 $$ $$ \cos 2\theta = 1 $$ Step 3: Solve the second condition for $\cos 2\theta$. From the second condition, we also isolate $\cos 2\theta$: $$ 2\cos 2\theta - 1 = 0 $$ $$ 2\cos 2\theta = 1 $$ $$ \cos 2\theta = \frac{1}{2} $$ Step 4: Analyze the simultaneity of the conditions. For the original equation to be an identity, both conditions derived in Step 1 must hold true for the *same* value of $\theta$. This implies that $\cos 2\theta$ must simultaneously satisfy: $$ \cos 2\theta = 1 \quad \text{and} \quad \cos 2\theta = \frac{1}{2} $$ These two requirements are contradictory because $\cos 2\theta$ cannot be equal to $1$ and $\frac{1}{2}$ at the same time. Step 5: Conclude based on the analysis. Since there is no value of $\cos 2\theta$ that satisfies both conditions simultaneously, there is no value of $\theta$ for which the given equation can be an identity. The final answer is $\boxed{1}$.