Consider the two quadratic equations is $x,$
$2 x^{2} + 4 x \cos θ + 5 \cos 2 θ = 3$ ,
$x^{2} - 2 x \cos θ + 2 \cos^{2} θ = 1$
If these two equations have atleast one common root for some $θ \in [0, 2 π]$ , then the number of possible value of $θ,$

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answer is 8.

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Detailed Solution

$\cos θ + \sin θ,$ $\cos θ - \sin θ$ are the roots of $x^{2} - 2 x \cos θ + 2 \cos^{2} θ = 1$
& $2 \sin θ - \cos θ, - 2 \sin θ - \cos θ$ are the roots of $2 x^{2} + 4 x \cos θ + 5 \cos 2 θ = 3$
$\cos θ + \sin θ = - 2 \sin θ - \cos θ \Rightarrow 2 \cos θ = - 3 \sin θ \Rightarrow$ 2 solutions
$\cos θ + \sin θ = 2 \sin θ - \cos θ \Rightarrow 2 \cos θ = \sin θ \Rightarrow$ 2 solutions
$\cos θ - \sin θ = 2 \sin θ - \cos θ \Rightarrow 2 \cos θ = 3 \sin θ \Rightarrow$ 2 solutions
$\cos θ - \sin θ = - 2 \sin θ - \cos θ \Rightarrow 2 \cos θ = - \sin θ \Rightarrow$ 2 solutions