Ответ:
t_1 = \frac{-8 + 12}{10} = 0,4; t _2 =\frac{-8-12}{10}=-2\\ cosx = t_1; arccos(x) = \{-1, 1\} = > cosx\neq t_2\\cosx=0.4=>x=arccos(0.4)+2\pi k" alt="5cos2x + 8cosx-4=0\\5(cos^2x-1) + 8cosx -4 = 0\\5cos^2x+8cosx - 9 = 0\\/cosx = t/\\5t^2+8t-9=0\\D = b^2 - 4ac = 64 + 180 => t_1 = \frac{-8 + 12}{10} = 0,4; t _2 =\frac{-8-12}{10}=-2\\ cosx = t_1; arccos(x) = \{-1, 1\} = > cosx\neq t_2\\cosx=0.4=>x=arccos(0.4)+2\pi k" align="absmiddle" class="latex-formula">