Решить такое уравнение. cos^x-sin^x+3sinx=2 . И найти точки принадлежащие отрезку. [pi;5pi/2]

а) cos^2x-sin^2x+3sinx-2=0 1-sin^2x-sin^2x+3sinx-2=0 -2sin^2x+3sinx-1=0 Пусть t=sinx, где t€[-1;1], тогда -2t^2+3t-1=0 D=9-8=1 t1=-3-1/-4=1 t2=-3+1/-4=-1/2 Вернёмся к замене: sinx=1 x=Π/2+2Πn, n€Z sinx=-1/2 x=-5Π/6+2Πk, k€Z x=-Π/6+2Πk, k€Z б) Решим с помощью двойного неравенства: 1) Π<=Π/2+2Πn<=5Π/2 Π-Π/2<=2Πn<=5Π/2-Π/2 Π/2<=2Πn<=4Π/2 Π/4<=Πn<=Π 1/4<=n<=1 n=1 x=Π/2+2Π*1=Π/2+2Π=5Π/2 2) Π<=-5Π/6+2Πk<=5Π/2 Π+5Π/6<=2Πk<=5Π/2+5Π/6 11Π/6<=2Πk<=20Π/6 11Π/12<=Πk<=20Π/12 11/12<=k<=20/12 k=1 x=-5Π/6+2Π*1=-5Π/6+2Π=7Π/6 3) Π<=-Π/6+2Πk<=5Π/2 Π+Π/6<=2Πk<=5Π/2+Π/6 7Π/6<=2Πk<=16Π/6 7Π/12<=Πk<=16Π/12 7/12<=k<=16/12 k=1 x=-Π/6+2Π*1=11Π/6 Ответ: а) Π/2+2Πn, n€Z; -5Π/6+2Πk, -Π/6+2Πk, k€Z; б) 7Π/6, 11Π/6, 5Π/2