Решить уравнение 6sin^2+cosx-5=0 и найдите корни,принадлежащие отрезку [2П,3П]

6(1 - cos^2x) + cosx - 5 = 0 6 - 6cos^2x + cosx - 5 = 0 1 - 6cos^2x + cosx = 0 6cos^2x - cosx - 1 = 0 Пусть cosx = t 6t^2 - t - 1 = 0 D = 5^2 t1 = 1/2 t2 = 1/3 cosx = 1/2 x = +-pi/3 + 2pin cosx = 1/3 x = +-(pi - arccos1/3) + 2pin [2pi ; 3pi] pi/3 + 2pin =gt; n = 1 =gt; 7pi/3 -pi/3 + 2pin =gt; -- (pi - arccos1/3) + 2pin =gt; n = 1 =gt; 3pi - arccos1/3 -(pi - arccos1/3) + 2pin =gt; --