<s>[INST] $\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$ what's the sum of possible $\theta$ $0 \leq \theta \leq 360$
$\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$
$\log_{2} 3 + \log_{2} \sin \theta = \log_{2} 9 + \log_{2} \cos ^2 \theta + \log_{2} 2$
$\log_{2} \sin \theta - \log_{2} \cos ^2 \theta = \log_{2} 6$
$\log_{2} \sin ^3 \theta = \log_{2} 6$ 
How to find the possible angle?
 [/INST] Clearly we need $\sin\theta>0,\cos\theta<0$
$$\implies90^\circ<\theta<180^\circ$$
Choose $180^\circ-\theta=t$
$$\log_2(3\sin t)=2\log_2(3\cos t)+1=\log_2(2(3\cos t)^2)$$
$$3\sin t=18(1-\sin ^2t)$$
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