0\; ,\; (x+2)^2>0\; \to \; \; x+2\ne 0\; ,\; \underline{x\ne -2}\\\\cos\frac{5\pi}{12}<1\; \; \to \; \; y=log_{cos\frac{5\pi }{12}}(x^2+4x+4)\; -\; ybuvaushaya\; \; \Rightarrow \\\\x^2+4x+4\leq 1\\\\(x+2)^2-1\leq 0\\\\(x+2-1)(x+2+1)\leq 0\\\\(x+1)(x+3)\leq 0\; \; ,\quad znaki:\; \; +++[-3\, ]---[-1\, ]+++\\\\x\in [-3,-1\, ]\\\\\left \{ {{x\ne -2} \atop {x\in [-3,-1\, ]}} \right. \; \; \Rightarrow \; \; \; \underline {\; x\in [-3,-2)\cup (-2,-1\, ]\; }" alt="log_{cos\frac{5\pi }{12}}\, (x^2+4x+4)\geq 0\; \; ,\\\\ODZ:\; x^2+4x+4>0\; ,\; (x+2)^2>0\; \to \; \; x+2\ne 0\; ,\; \underline{x\ne -2}\\\\cos\frac{5\pi}{12}<1\; \; \to \; \; y=log_{cos\frac{5\pi }{12}}(x^2+4x+4)\; -\; ybuvaushaya\; \; \Rightarrow \\\\x^2+4x+4\leq 1\\\\(x+2)^2-1\leq 0\\\\(x+2-1)(x+2+1)\leq 0\\\\(x+1)(x+3)\leq 0\; \; ,\quad znaki:\; \; +++[-3\, ]---[-1\, ]+++\\\\x\in [-3,-1\, ]\\\\\left \{ {{x\ne -2} \atop {x\in [-3,-1\, ]}} \right. \; \; \Rightarrow \; \; \; \underline {\; x\in [-3,-2)\cup (-2,-1\, ]\; }" align="absmiddle" class="latex-formula">