а)корень из 2 на cos^2=sin(x-pi/2)
б) отметить точки на промежутке от [-3п/2;-п]

√2cos²x+cosx=0cosx(√2cosx+1)=0cosx=0x=π/2+πn,n∈z-3π/2≤π/2+πn≤-π-3≤1+2n≤-2-4≤2n≤-3-2≤n≤-1,5n=-2⇒x=π/2-2π=-3π/2√2cosx+1=0cosx=-1/√2x=-3π/4+2πk,k∈z U x=3π/4+2πt,t∈z-3π/2≤-3π/4+2πk≤-π-6≤-3+8k≤-4-3≤8k≤-1-3/8≤k≤-1/8нет решения-3π/2≤3π/4+2πt≤-π-6≤3+8t≤-4-9≤8t≤-7-9/8≤t≤-7/8t=-1⇒x=3π/4-2π=-5π/4