Определите количество корней уравнения 6sin²x+2sin²2x=5 на промежутке [π/2; π]

6*(1-cos2x)/2+2*(1-cos4x)/2=5
3-3cos2x+1-cos4x-5=0
cos4x+3cos2x+1=0
2cos²2x-1+3cos2x+1=0
2cos²2x+3cos2x=0
cos2x(2cos2x+3)=0
cos2x=0⇒2x=π/2+πn,n∈z⇒x=π/4+πn/2,n∈z
π/2≤π/4+πn/2≤π
2≤1+2n≤4
1≤2n≤3
1/2≤n≤3/2
n=1⇒x=π/4+π/2=3π/4
2cos2x+3=0⇒cos2x=-1,5<-1 нет решения