А)1-2sin^2x+3sin^2x=5/4
-2sin^2x+3sin^2x=1/4
sin^2x=1/4
sinx=1/2
x1=Π/6+2Πk, k€Z
x2=5Π/6+2Πn, n€Z
б) решим с помощью двойного неравенства:
Π<=Π/6+2Πk<=5Π/2<br>5Π/6<=2Πk<=14Π/6<br>5Π/12<=Πk<=14Π/12<br>5/12<=k<=14/12<br>k=1
Π/6+2Π*1=13Π/6 - первый корень
Π<=5Π/6+2Πn<=5Π/2<br>Π/6<=2Πn<=10Π/6<br>Π/12<=Πn<=10Π/12<br>1/12<=n<=10/12 не имеет корней <br>Ответ: а) Π/6+2Πk,k€Z; 5Π/6+2Πn, n€Z, б) 13Π/6