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

А) cos^2x-sin^2x+3sinx-2=01-sin^2x-sin^2x+3sinx-2=0-2sin^2x+3sinx-1=0Пусть t=sinx, где t€[-1;1], тогда-2t^2+3t-1=0D=9-8=1t1=-3-1/-4=1t2=-3+1/-4=-1/2Вернёмся к замене:sinx=1x=Π/2+2Πn, n€Zsinx=-1/2x=-5Π/6+2Πk, k€Zx=-Π/6+2Πk, k€Zб) Решим с помощью двойного неравенства:1) Π<=Π/2+2Πn<=5Π/2Π-Π/2<=2Πn<=5Π/2-Π/2Π/2<=2Πn<=4Π/2Π/4<=Πn<=Π1/4<=n<=1n=1x=Π/2+2Π*1=Π/2+2Π=5Π/22) Π<=-5Π/6+2Πk<=5Π/2Π+5Π/6<=2Πk<=5Π/2+5Π/611Π/6<=2Πk<=20Π/611Π/12<=Πk<=20Π/1211/12<=k<=20/12k=1x=-5Π/6+2Π*1=-5Π/6+2Π=7Π/63) Π<=-Π/6+2Πk<=5Π/2Π+Π/6<=2Πk<=5Π/2+Π/67Π/6<=2Πk<=16Π/67Π/12<=Πk<=16Π/127/12<=k<=16/12k=1x=-Π/6+2Π*1=11Π/6Ответ: а) Π/2+2Πn, n€Z; -5Π/6+2Πk, -Π/6+2Πk, k€Z; б) 7Π/6, 11Π/6, 5Π/2