Question

cos 2x-7cosx+4= 0 и cos2x не в квадрате! Решите уравнение

Answer 1

$cos2x-7cosx+4=0 \(2cos^2x-1)-7cosx+4=0 \cosx=t, t in [-1;1] \2t^2-7t+3=0 \D=49-24=25=5^2 \t_1= frac{7+5}{4} notin [-1;1] \t_2= frac{7-5}{4}= frac{1}{2} \cosx=frac{1}{2} \x= pm frac{pi}{3} +2pi n, n in Z$