Log 13 (cos2x -9корня из 2 cosx -8)=0

$log_{13}(cos2x -9 \sqrt{2} \ cosx -8)=0$
$log_{13}(cos2x -9 \sqrt{2} \ cosx -8)=log1$
$cos2x -9 \sqrt{2} \ cosx -8=1$
$cos2x -9 \sqrt{2} \ cosx -8-1=0$
$cos2x -9 \sqrt{2} \ cosx -9=0$
$2cos^2x-1 -9 \sqrt{2} \ cosx -9=0$
$2cos^2x -9 \sqrt{2} \ cosx -10=0$
Замена: $cosx=a,$ $|a| \leq 1$
$2a^2-9 \sqrt{2} a-10=0$
$D=(-9 \sqrt{2})^2-4*2*(-10)=242$
$a_1= \frac{9 \sqrt{2} -11 \sqrt{2} }{4} =- \frac{ \sqrt{2} } {2}$
$a_2= \frac{9 \sqrt{2} +11 \sqrt{2} }{4} =5 \sqrt{2}$ ∅

$cosx=- \frac{ \sqrt{2} }{2}$
$x=бarccos( - \frac{ \sqrt{2} }{2}) +2\pi n,$ $n$ ∈ $Z$
$x=б( \pi -arccos \frac{ \sqrt{2} }{2} )+2 \pi n,$ $n$ ∈ $Z$
$x=б( \pi - \frac{ \pi }{4} )+2 \pi n,$ $n$ ∈ $Z$
$x=б \frac{ 3 \pi }{4} +2 \pi n,$ $n$ ∈ $Z$

Проверка:
$log_{13}(cos(2* \frac{3 \pi }{4} ) -9 \sqrt{2} \ cos \frac{3 \pi }{4} -8)=0$
$log_{13}(cos \frac{3 \pi }{2} -9 \sqrt{2} \ cos( \pi - \frac{ \pi }{4}) -8)=0$
$log_{13}(9 \sqrt{2}* \frac{ \sqrt{2} }{2} -8)=0$
$log_{13}1=0$

$log_{13}(cos(2* (-\frac{3 \pi }{4} )) -9 \sqrt{2} \ cos(- \frac{3 \pi }{4} )-8)=0$
$log_{13}(cos (-\frac{3 \pi }{2})-9 \sqrt{2} \ cos( \pi - \frac{ \pi }{4}) -8)=0$
$log_{13}(9 \sqrt{2}* \frac{ \sqrt{2} }{2} -8)=0$
$log_{13}1=0$

Ответ: $x=б \frac{ 3 \pi }{4} +2 \pi n,$ $n$ ∈ $Z$