Question

Ребята,ОЧЕНЬ СРОЧНО НУЖНО РЕШИТЬ!!! АЛГЕБРА

Answer 1

а)

$9^{\cos x}=9^{\sin x}\cdot 3^{\frac{2}{\cos x}}\medskip\\9^{\cos x}=9^\sin x}\cdot 9^{\frac{2}{2\cos x}}\medskip\\9^{\cos x}=9^{\sin x+\frac{1}{\cos x}}\medskip\\\cos x=\sin x+\dfrac{1}{\cos x}\medskip\\1)~D(y):\cos x\neq 0\medskip\\D(y): x\neq\dfrac{\pi}{2}+\pi n,~n\in\mathbb{Z}\medskip\\2)~\cos^2 x=\cos x\sin x+1\medskip\\1-\sin^2 x=\cos x\sin x+1\medskip\\\cos x\sin x+\sin^2 x=0\medskip\\\sin x\left(\cos x+\sin x\right)=0\medskip\\\left[\begin{gathered}\sin x=0\\\cos x+\sin x=0\end{gathered}$

$\left[\begin{gathered}x=\pi m,~m\in\mathbb{Z}\\\mathrm{tg}~x=-1\end{gathered}\Leftrightarrow \left[\begin{gathered}x=\pi m,~m\in\mathbb{Z}\\x=-\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}\end{gathered}$

$x\in\left\{\pi m,~m\in\mathbb{Z}\right\}\cup\left\{-\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}\right\}$

б)

$\left[-\dfrac{7\pi}{2};-\dfrac{5\pi}{2}\right]\medskip\\1)~-3{,}5\pi\leqslant \pi m\leqslant -2{,}5\pi\medskip\\-3{,}5\leqslant m\leqslant -2{,}5\medskip\\m\in\mathbb{Z}\Rightarrow m\in\left\{-3\right\}\medskip\\x_1=-3\pi\medskip\\2)~-3{,}5\pi\leqslant -\dfrac{\pi}{4}+\pi k\leqslant -2{,}5\pi\medskip\\-3{,}5\leqslant -\dfrac{1}{4}+k\leqslant -2{,}5\medskip\\-3{,}25\leqslant k\leqslant -2{,}25\medskip\\k\in\mathbb{Z}\Rightarrow k\in\left\{-3\right\}$

$x_2=-\dfrac{\pi}{4}-3\pi=-\dfrac{13\pi}{4}$

Ответ. а) $x\in\left\{\pi m,~m\in\mathbb{Z}\right\}\cup\left\{-\dfrac{\pi}{4}+\pi k,~k\in\mathbb{Z}\right\}$; б) $\left\{-3\pi;-\dfrac{13\pi}{4}\right\}$