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(D) The given equation is of the form $a \sin x + b \cos x = c$, where $a = 2$, $b = 1$, and $c = 3$.For the equation $a \sin x + b \cos x = c$ to hav

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(D) The given equation is of the form $a \sin x + b \cos x = c$, where $a = 2$, $b = 1$, and $c = 3$.For the equation $a \sin x + b \cos x = c$ to have a solution, the condition $|c| \leq \sqrt{a^2 + b^2}$ must be satisfied.Here, $\sqrt{a^2 + b^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.Since $\sqrt{5} \approx 2.236$, we have $c = 3 > \sqrt{5}$.Because the maximum value of $2 \sin x + \cos x$ is $\sqrt{5}$, it can never reach the value $3$.Therefore, the equation has no solution.

The number of solutions of $2 \sin x + \cos x = 3$ is

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