а) Решим уравнение

$\frac{3^{\cos x}}{9^{\cos^{2}x}}=\frac{3^{\cos x}}{3^{2\cos^{2}x}}=3^{\cos x-2\cos^{2}x}$ $3^{\cos x-2\cos^{2}x}=4^{2\cos^{2}x-\cos x}\Leftrightarrow\left(\frac{1}{3}\right)^{2\cos^{2}x-\cos x}=4^{2\cos^{2}x-\cos x}\Leftrightarrow\left(\frac{1}{12}\right)^{2\cos^{2}x-\cos x}=1\Leftrightarrow$ $\Leftrightarrow2\cos^{2}x-\cos x=0\Leftrightarrow\left[\begin{matrix}{\cos x=0}\\{\cos x=\frac{1}{2}}\\\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}{x=\frac{\pi}{2}+\pi k}\\{x=-\frac{\pi}{3}+2\pi k ,k\in\mathbb{Z}}\\{x=\frac{\pi}{3}+2\pi k}\\\end{matrix}\right.$

б) Отберем корни с помощью двойного неравенства:

$\begin{gathered} -\frac{3\pi}{2}\leq\frac{\pi}{2}+\pi k\leq\frac{\pi}{6}\mid:\pi \\ -\frac{3}{2}\leq\frac{1}{2}+k\leq\frac{1}{6}\mid-\frac{1}{2} \\ -\frac42\leq k\leq-\frac26 \\ -\frac{12}{6}\leq k\leq-\frac{2}{6} \\ k\in\mathbb{Z}\Rightarrow k=-2;-1\Rightarrow \\ \Rightarrow x_{1}=\frac{\pi}{2}-2\pi=-\frac{3\pi}{2};x_{2}=\frac{\pi}{2}-\pi=-\frac{\pi}{2} \end{gathered}$

$\begin{gathered} -\frac{3\pi}{2}\leq-\frac{\pi}{3}+2\pi k\leq\frac{\pi}{6}\mid:\pi \\ -\frac{3}{2}\leq-\frac{1}{3}+2k\leq\frac{1}{6}\mid+\frac{1}{3} \\ -\frac{7}{6}\leq2k\leq\frac{3}{6}\mid:2 \\ -\frac{7}{12}\leq k\leq\frac{3}{12} \\ k\in\mathbb{Z}\Rightarrow k=0\Rightarrow \\ \Rightarrow x_{3}=-\frac{\pi}{3} \end{gathered}$

$\begin{gathered} -\frac{3\pi}{2}\leq\frac{\pi}{3}+2\pi k\leq\frac{\pi}{6}\mid:\pi \\ -\frac{3}{2}\leq\frac{1}{3}+2k\leq\frac{1}{6}\mid-\frac{1}{3} \\ -\frac{11}{6}\leq2k\leq-\frac{1}{6}\mid:2 \\ -\frac{11}{12}\leq k\leq-\frac{1}{12} \\ k\in\mathbb{Z}\Rightarrow k\in\emptyset \end{gathered}$

Задания для подготовки к ЕГЭ Уравнения смешанного типа