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Solve: 2sinθ(sin2θ+sin4θ+sin6θ+⋯+sin14θ)=cosθ−122\sin\theta(\sin2\theta +\sin4\theta +\sin6\theta +\dots+\sin14\theta ) = \cos\theta - \frac122sinθ(sin2θ+sin4θ+sin6θ+⋯+sin14θ)=cosθ−21 where 0∘≤θ≤24∘0^{\circ}\leq \theta \leq 24^{\circ}0∘≤θ≤24∘.
Give your answer as the product of your solutions (each in degrees).
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