[tex]\log_2(2\sin x)+\log_2(\cos x)=-1[/tex]
Condense the logarithms on the left side:
[tex]\log_2(2\sin x\cos x)=-1[/tex]
Recall the double angle identity for sine, [tex]\sin(2x)=2\sin x\cos x[/tex]:
[tex]\log_2(\sin(2x))=-1[/tex]
Write both sides as powers of 2:
[tex]2^{\log_2(\sin(2x))}=2^{-1}[/tex]
Simplify this:
[tex]\sin(2x)=\dfrac12[/tex]
Solve for [tex]2x[/tex]:
[tex]2x=\sin^{-1}\left(\dfrac12\right)+2n\pi\text{ OR }2x=\pi-\sin^{-1}\left(\dfrac12\right)+2n\pi[/tex]
(where [tex]n[/tex] is any integer)
Recall that [tex]\sin^{-1}\left(\frac12\right)=\frac\pi6[/tex]:
[tex]2x=\dfrac\pi6+2n\pi\text{ OR }2x=\dfrac{5\pi}6+2n\pi[/tex]
Solve for [tex]x[/tex]:
[tex]x=\dfrac\pi{12}+n\pi\text{ OR }x=\dfrac{5\pi}{12}+n\pi[/tex]
We get solutions in the interval [tex]2\pi <x<\frac{5\pi}2[/tex] when [tex]n=2[/tex], giving
[tex]\boxed{x=\dfrac{25\pi}{12}}\text{ OR }\boxed{x=\dfrac{29\pi}{12}}[/tex]
