How many solutions does this equation involving $\sin^8x$ have?

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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).