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For $$0 < \theta < \frac{\pi}{2}$$, the solution(s) of$$ \sum_{m=1}^{6}\cosec\left(\theta + \frac{(m - 1)\pi}{4}\right)\cosec\left(\theta + \frac{m\pi}{4}\right) = 4\sqrt{2}$$is(are)
$$\frac{\pi}{4}$$
$$\frac{\pi}{6}$$
$$\frac{\pi}{12}$$
$$\frac{5 \pi}{12}$$
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