log-2-sin-x-5pi-12-log-2-sin-x-pi-12-1-

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

\log_{2} (\sin (x + \frac{5 π}{12})) + \log_{2} (\sin (x + \frac{π}{12})) = - 1

Commented by jagoll last updated on 20/Apr/20

\Rightarrow \sin (x + \frac{5 π}{12}) \sin (x + \frac{π}{12}) = \frac{1}{2}

\cos (\frac{4 π}{12}) - \cos (2 x + \frac{6 π}{12}) = 1

\frac{1}{2} - \cos (2 x + \frac{π}{2}) = 1

\frac{1}{2} + \sin 2 x = 1 \Rightarrow \sin 2 x = \frac{1}{2}

{\begin{cases} 2 x = \frac{π}{6} + 2 n π \\ 2 x = \frac{5 π}{6} + 2 n π \end{cases}

{\begin{cases} x = \frac{π}{12} + n π \\ x = \frac{5 π}{12} + n π \end{cases}

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