Решите тригонометрию. 2.

Question 1

Question 2

$\cos{5x}*\cos{4x}+\cos{4x}*\cos{3x}=\cos^2{2x}*\cos{x}\\\cos{4x}(\cos{5x}+\cos{3x})-\cos^2{2x}*\cos{x}=0\\\cos{4x}*2\cos{4x}*\cos{x}-\cos^2{2x}*\cos{x}=0\\\cos{x}(2\cos^2{4x}-\cos^2{2x})=0\\\begin{bmatrix}\cos{x}=0\\\begin{matrix}2\cos^2{4x}-\cos^2{2x}=0&(1)\end{matrix}\end{matrix}\\(1)2\cos^2{4x}-\frac{1+\cos{4x}}{2}=0|*2\\4\cos^2{4x}-\cos{4x}-1=0;D=1+16$

-9\\\cos{4x}=\frac{1+\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}\sqrt{17}<7\end{matrix}\\Otvet:x=\begin{Bmatrix}\pm \frac{1}{4}\arccos{(\frac{1\pm \sqrt{17}}{8})}+\frac{\pi n}{2};\pm \frac{\pi }{2}+2\pi n\end{Bmatrix},n\in Z." alt="\begin{bmatrix}\cos{4x}=\frac{1-\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}-\sqrt{17}>-9\\\cos{4x}=\frac{1+\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}\sqrt{17}<7\end{matrix}\\Otvet:x=\begin{Bmatrix}\pm \frac{1}{4}\arccos{(\frac{1\pm \sqrt{17}}{8})}+\frac{\pi n}{2};\pm \frac{\pi }{2}+2\pi n\end{Bmatrix},n\in Z." align="absmiddle" class="latex-formula">

WhatYouNeed_zn · Answer 1 · 2020-06-03T15:19:18+0000

$\cos{5x}*\cos{4x}+\cos{4x}*\cos{3x}=\cos^2{2x}*\cos{x}\\\cos{4x}(\cos{5x}+\cos{3x})-\cos^2{2x}*\cos{x}=0\\\cos{4x}*2\cos{4x}*\cos{x}-\cos^2{2x}*\cos{x}=0\\\cos{x}(2\cos^2{4x}-\cos^2{2x})=0\\\begin{bmatrix}\cos{x}=0\\\begin{matrix}2\cos^2{4x}-\cos^2{2x}=0&(1)\end{matrix}\end{matrix}\\(1)2\cos^2{4x}-\frac{1+\cos{4x}}{2}=0|*2\\4\cos^2{4x}-\cos{4x}-1=0;D=1+16$

-9\\\cos{4x}=\frac{1+\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}\sqrt{17}<7\end{matrix}\\Otvet:x=\begin{Bmatrix}\pm \frac{1}{4}\arccos{(\frac{1\pm \sqrt{17}}{8})}+\frac{\pi n}{2};\pm \frac{\pi }{2}+2\pi n\end{Bmatrix},n\in Z." alt="\begin{bmatrix}\cos{4x}=\frac{1-\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}-\sqrt{17}>-9\\\cos{4x}=\frac{1+\sqrt{17}}{8}\begin{vmatrix}\\\\\end{matrix}\sqrt{17}<7\end{matrix}\\Otvet:x=\begin{Bmatrix}\pm \frac{1}{4}\arccos{(\frac{1\pm \sqrt{17}}{8})}+\frac{\pi n}{2};\pm \frac{\pi }{2}+2\pi n\end{Bmatrix},n\in Z." align="absmiddle" class="latex-formula">