-5cos4x = 2cos(^2)x + 1Ответ: +- pi/6 +pi*k; +-1/2arccos(-3/5) +pi*kНужно решение как...

Решите задачу:

$-5cos4x=2cos^2x+1\\\\\star \; \; cos^2x= \frac{1+cos2x}{2} \; \; \Rightarrow \; \; 2cos^2x=1+cos2x\; \; \; \star \\\\-5cos4x=(1+cos2x)+1\\\\cos2x+5cos4x+2=0\\\\\star \; \; 1+cos4x=2cos^22x\; \; \Rightarrow \; \; cos4x=2cos^22x-1\; \; \star \\\\co2x+5(2cos^22x-1)+2=0\\\\10cos^22x+cos2x-3=0\\\\D=1+120=121\; ,\\\\a)\quad (cos2x)_1= \frac{-1-11}{20} =-\frac{12}{20}=-\frac{3}{5}\\\\2x=\pm arccos(-\frac{3}{5})+2\pi n,\; n\in Z\\\\x=\pm \frac{1}{2}(\pi -arccos\frac{3}{5})+\pi n\; ,\; n\in Z$

$b)\quad (cos2x)_2=\frac{-1+11}{20}=\frac{10}{20}=\frac{1}{2}$

$2x=\pm \frac{\pi}{3}+2\pi m\; ,\; m\in Z\\\\x=\pm \frac{\pi}{6}+\pi m\; ,m\in Z$