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 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