Question

Решить на егэ вроде решил, но не знаю правильно ли, хочу проверить: cos2x-5√2cosx-5=0

Answer 1

$\displaystyle \sf \cos 2x-5 \sqrt{2} \cos x-5=0\\ (2\cos^2x-1)-5\sqrt{2}\cos x-5=0\\2\cos^2x-5\sqrt{2}\cos x-6=0\\ \cos x=t; \;\;\;\; t\in[-1;1]\\2t^2-5\sqrt{2}t-6=0\\ D=(5\sqrt{2})^2-4\cdot 2\cdot (-6)=50+48=98=(7\sqrt{2})^2, D0 \\\\t_{1,2} = \frac{5\sqrt{2}\pm 7\sqrt{2}}{4}=\left |{ {{3\sqrt{2} } \atop {-\frac{\sqrt{2}}{2}}} \right. \\\\ t=-\frac{\sqrt{2}}{2}\\ \cos x=-\frac{\sqrt{2}}{2}\\ \left [{ {{x=\frac{3\pi}{4}+2\pi n, n\in Z} \atop {x=-\frac{3\pi }{4}+2\pi l, l\in Z}} \right.$

ответ: $\displaystyle \sf -\frac{3\pi}{4}+2\pi n, n\in Z; \frac{3\pi }{4}+2\pi l, l\in Z.$