а) cos2x+1=√2cos(П/2) б) на промежутке [2П;7П/2] плизз решите подробно на листочке)))

2сosx+1=0cosx=-1/2x=-2π/3+2πn U x=2π/3+2πk2π≤-2π/3+2πn≤7π/212≤-4+12n≤2116≤12n≤254/3≤n≤25/12n=2⇒x=-2π/3+4π=10π/32π≤2π/3+2πk≤7π/212≤4+12k≤218≤12k≤172/3≤k≤17/12k=1⇒x=2π/3+2π=8π/3