In the range \(0 \leq x < 2\pi\) the equation \[(3 + \cos x)^2 = 4 - 2\sin^8x\] has

\(0\) solutions,

\(1\) solution,

\(2\) solutions,

\(3\) solutions.

The quickest way to see this is to notice that \[(3+\cos x)^2 \geq (3-1)^2 = 4\] and \[4 - 2\sin^8x \leq 4.\] So there can only be solutions when these hold with equality, i.e. when \(\cos x = -1\) and \(\sin x = 0\). In the range \(0 \leq x < 2\pi\) this only occurs at \(x = \pi\), so the answer is (b).