Show that \(3\cos{2x} + 11\sin{x} = 3 + 11\sin{x} - 6\sin^2{x}\).
Hence, or otherwise, solve the equation \(3\cos{2x} + 11\sin{x} - 6 = 0\) for \(0^{\circ} \leq x \leq 180^{\circ}\).
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