If 2sin3x+sin2xcosx+4sinx−4=0 has exactly 3 solutions in the interval [0,2nπ], n∈N, then the roots of the equation x2+nx+(n−3)=0 belong to:
Solutions · 1
Official solution
*(The source PDF supplies the stem, metadata, options, and answer key letter B, but the worked solution text was not present in the extraction, and attempting to re-derive the exact equation independently did not reproduce a clean value of n consistent with verification --- so the derivation is not reproduced here to avoid presenting an unverified chain of reasoning as fact. The answer B, i.e. that the roots of x2+nx+(n−3)=0 lie in (−∞,0), is taken directly from the source's answer key.)*