Question

Алгебра. Пожалуйста. Срочно
Много баллов даю) ​

Answer 1

$(81^{sinx})^{cosx}=9^{\sqrt{2}cosx}\\(9^{2sinx})^{cosx}=9^{\sqrt{2}cosx}\\2sinxcosx=\sqrt{2}cosx\\2sinxcosx-\sqrt{2}cosx=0\\cosx(2sinx-\sqrt{2})=0\\\left[\begin{gathered}cosx=0,\\sinx=\frac{\sqrt{2}}{2}\end{gathered}\right.\\\left[ \begin{gathered}x= \frac{\pi }{2} +\pi k,k\in \mathbb{Z},\\x=\frac{\pi }{4} +2\pi n, n\in \mathbb{Z}\end{gathered}\right.$

$\frac{\pi }{2} \leq \frac{\pi }{2} +\pi k\leq 2\pi \\ 0\leq k\leq 1,5\\ k=0;1 \Rightarrow x=\frac{\pi }{2} +\pi =\frac{3\pi }{2}\; or\; x=\frac{\pi }{2}$

$\frac{\pi }{2} \leq \frac{\pi }{4} +2\pi n\leq 2\pi\\\frac{1}8\leq n \leq \frac{7}{8}$