Answer:
[tex]$x=\frac{25\pi }{12} \text{ and }x=\frac{29\pi }{12}[/tex]
Step-by-step explanation:
Solve
[tex]\log_2(2 \sin x) + \log_2(\cos x) =1[/tex]
in interval [tex]$\left(2\pi, \frac{5\pi}{2} \right)$[/tex]
Remember that logarithms follow:
[tex]f(x\cdot y)=f(x)+ f(y)[/tex]
Therefore
[tex]\log_2(2 \sin x) + \log_2(\cos x) =-1 \implies \log_2(2 \sin x \cdot \cos x) =-1[/tex]
We got a double angle of sine:
[tex]\sin(2x)= 2 \sin x \cos x[/tex]
[tex]\log_2( \sin 2x ) =-1[/tex]
[tex]2^{-1}=\sin 2x[/tex]
[tex]$\frac{1}{2} =\sin 2x$[/tex]
The solutions for it are:
[tex]$2x=\frac{\pi }{6}+2\pi n \text{ and }2x=\frac{5\pi }{6}+2\pi n, \text{ as } n\in \mathbb{Z}$[/tex]
As
[tex]$x=\frac{\pi }{12}+\pi n \text{ and }x=\frac{5\pi }{12}+\pi n, \text{ as } n\in \mathbb{Z}$[/tex]
But we have the interval
[tex]$\left(2\pi, \frac{5\pi}{2} \right)$[/tex]
Therefore, the answer is
[tex]$x=\frac{25\pi }{12} \text{ and }x=\frac{29\pi }{12}[/tex]
