Question

Помогите решить тринадцатый номер, пожалуйста!

Answer 1

$\sqrt{(4cos3x-5)^2}-\sqrt{cos^23x-8cos3x+16}=4\\\\\sqrt{(4cos3x-5)^2}-\sqrt{(cos3x-4)^2}=4\\\\|4cos3x-5|-|cos3x-4|=4\\\\t=cos3x,\; \; |4t-5|-|t-4|=4\\\\4t-5=0\; \to \; t=\frac{5}{4}=1,25\\\\t-4=0\; \to \; t=4\\\\Znaki\; (4t-5):\; \; ---(\frac{5}{4})+++(4)+++\\\\Znaki\; (t-4):\; \; ---(\frac{5}{4})---(4)+++$

$1)\; t \leq \frac{5}{4}\; \; \to \; |4t-5|=-(4t-5)=5-4t,\; \; |t-4|=-(t-4)=4-t\\\\5-4t-(4-t)=4\; ,\; \; -3t=3,\; \; t=-1\\\\cos3x=-1\; ,\; \; 3x=\pi +2\pi n,\; \; \; x=\frac{\pi}{3}+\frac{2\pi n}{3},\; n\in Z\\\\2)\; \; \frac{5}{4}\ \textless \ t \leq 4\; \; \to \; |4t-5|=4t-5,\; \; |t-4|=-(t-4)=4-t\\\\4t-5-(4-t)=4\; ,\; \; 5t=13\; ,\; \; t=\frac{13}{5}\ \textgreater \ 1\; \to \\\\cos3x=\frac{13}{5}\; \; ne\; imeet\; reshenij\\\\3)\; \; t\ \textgreater \ 4\; \to \; |4t-5|=4t-5\; ,\; \; |t-4|=t-4\\\\4t-5-(t-4)=4\; ,\; \; 3t=5,$

$t=\frac{5}{3}\notin (4,+\infty )$

Oтвет: $x=\frac{\pi}{3}+\frac{2\pi}{3}n,n\in Z$