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Question Number 89951 by john santu last updated on 20/Apr/20

$$\log {}_2\left(\sin (x+\frac{5\pi }{12})\right)+\log {}_2\left(\sin (x+\frac{\pi }{12})\right)=-1$$

Commented by jagoll last updated on 20/Apr/20

$$\Rightarrow \sin (\mathrm{x}+\frac{5\pi }{12})\sin (\mathrm{x}+\frac{\pi }{12})=\frac{1}{2}$$$$\cos \left(\frac{4\pi }{12}\right)-\cos (2\mathrm{x}+\frac{6\pi }{12})=1$$$$\frac{1}{2}-\cos (2\mathrm{x}+\frac{\pi }{2})=1$$$$\frac{1}{2}+\sin 2\mathrm{x}=1\Rightarrow \sin 2\mathrm{x}=\frac{1}{2}$$$$\{\begin{array}{l}2\mathrm{x}=\frac{\pi }{6}+2\mathrm{n}\pi \\ 2\mathrm{x}=\frac{5\pi }{6}+2\mathrm{n}\pi \end{array}$$$$\{\begin{array}{l}\mathrm{x}=\frac{\pi }{12}+\mathrm{n}\pi \\ \mathrm{x}=\frac{5\pi }{12}+\mathrm{n}\pi \end{array}$$$$$$

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