СРОЧНО!!! РЕШИТ ПОЖАЛУЙСТА ПОКАЗАТЕЛЬНОЕ НЕРАВЕНСТВО И ЛОГАРИМФМИЧЕСКОЕ УРАВНЕНИЕ. !!!! С ПОЛНЫМ РЕШЕНИЕМ И ОБОСНОВАНИЕМ. #13, #15. ДАЮ 14 БАЛЛОВ.

[latex]1)\quad lg^2(tg^2x)+lg(cosx)=lg(sinx)\; ,\quad x\in [\frac{\pi}{3},2\pi]\\\\\star \; \; lg^2(tg^2x)=(lg(tg^2x))^2=\left (lg\frac{sin^2x}{cos^2x}\right )^2=\\\\=\left (lg(sinx)^2-lg(cosx)^2)\right )^2=(2lg(sinx)-2lg(cosx))^2=\\\\=4(lg(sinx)-lg(cosx))^2\; \star \\\\Oboznachim:\; \; a=lg(sinx)\; ,\; \; b=lg(cosx)\\\\4(a-b)^2+b=a\\\\4(a-b)^2+b-a=0\\\\4(a-b)^2-(a-b)=0\\\\(a-b)(4(a-b)-1)=0\\\\(a-b)(4a-4b-1)=0\\\\a-b=0\; \; \; ili\; \; \; 4a-4b-1=0\\\\a)\; \; lg(sinx)-lg(cosx)=0[/latex] [latex]lg(sinx)=lg(cosx)\; \; \; \to \; \; \; sinx=cosx\; |:cosx\ne 0\\\\tgx=1\; \; \to \; \; \underline {x=\frac{\pi}{4}+\pi n,\; n\in Z}\\\\b)\; \; 4lg(sinx)-4lg(cosx)-1=0\\\\4\cdot lg\left (\frac{sinx}{cosx}\right )=1\; \; \to \; \; lg\left (\frac{sinx}{cosx}\right )^4=lg10\; \; \to \; \; lg(tgx)^4=lg10\\\\tg^4x=10\; \; \to \; \; tgx=\pm \sqrt[4]{10}\; \; \to \; \; \underline {x=\pm arctg\sqrt[4]{10}+\pi k,\; k\in Z}\\\\c)\; \; x\in [\frac{\pi}{3};2\pi ]\; \; \to \; \; x=\frac{\pi}{4}+\pi =\frac{5\pi}{4}\; ;[/latex] [latex]x=arctg\sqrt[4]{10}+\pi ,\; t.k.\; \; arctg\frac{\pi}{3}=\sqrt3\approx 1,732,\; a\; \sqrt[4]{10}\approx 1,778\\\\x=-arctg\sqrt[4]{10}\\\\Otvet:\; \; \frac{5\pi}{4}\; ,\; \; arctg\sqrt[4]{10}+\pi \; ,\; x=-arctg\sqrt[4]{10}.[/latex] [latex]2)\quad \frac{64^{x}-7\cdot 16^{x}}{4^{x}+1} + \frac{6\cdot 16^{x}-3\cdot 4^{x+2}+42}{4^{x}-6} \geq 0\\\\\frac{4^{3x}-7\cdot 4^{2x}}{4^{x}+1} + \frac{6\cdot 4^{2x}-3\cdot 4^2\cdot 4^{x}+42}{4^{x}-6} \geq 0\\\\t=4^{x}\ \textgreater \ 0\; :\quad \frac{t^3-7t^2 }{t+1} + \frac{6t^2-48t+42}{t-6}\geq 0\\\\6t^2-48t+42=6\cdot \underbrace {(t^2-8t+7)}_{t_1=1,t_2=7}=6\cdot (t-1)(t-7)\\\\ \frac{t^2(t-7)}{t+1} + \frac{6(t-1)(t-7)}{t-6} \geq 0\; ,\; \; \frac{t^2(t-7)(t-6)+6(t-1)(t-7)(t+1)}{(t+1)(t-6)} \geq 0[/latex] [latex]\frac{(t-7)(t^2(t-6)+6(t-1)(t+1))}{(t+1)(t-6)}\geq 0\\\\\frac{(t-7)(t^3-6t^2+6t^2-6)}{(t+1)(t-6)}\geq 0\; \ ;\; \; \frac{(t-7)(t^3-6)}{(t+1)(t-6)} \geq 0[/latex] [latex]\frac{(t-7)(t-\sqrt[3]6)(t^2+\sqrt[3]6\cdot t+\sqrt[3]{36})}{(t+1)(t-6)} \geq 0[/latex]  [latex]t^2+\sqrt[3]6t+\sqrt[3]{36}\ \textgreater \ 0\; ,\; t.k.[/latex] [latex]D=\sqrt[3]{36}-4\cdot \sqrt[3]{36}=-3\sqrt[3]{36}<0;[\tex] [latex] \sqrt[3]{6}\approx 1,817 [/latex] [latex]+++(-1)---[\sqrt[3]6\, ]+++(6)---[\, 7\, ]+++[/latex] [latex]t\in (-\infty ,-1)\cup [\sqrt[3]6,\, 6)\cup [\, 7,+\infty )[/latex] [latex]t=4^{x}\ \textgreater \ 0\; \; \to \; \; \sqrt[3]6\ \textless \ 4^{x}\ \textless \ 6\; ,\; 4^{x}\ \textgreater \ 7\\\\4^{log_4\sqrt[3]6}\ \textless \ 4^{x}\ \textless \ 4^{log_46}\; ,\; \; 4^{x}\ \textgreater \ 4^{log_47}\\\\log_4\sqrt[3]6\ \textless \ x\ \textless \ 6\; ,\; \; x\ \textgreater \ log_47\\\\x\in (log_4\sqrt[3]6\; ;\; 6)\cup (log_47\; ;+\infty )[/latex]