Question

cos(2x-19П/2)=sin(П/2+x)
отобрать корни на промежутке [-7П/2;-5П/2]

Answer 1

-sin2x=cosx
cosx+2sinxcosx=0
cosx(1+2sinx)=0
cosx=0⇒x=π/2+πn
sinx=-1/2⇒x=(-1)^n+1 *π/6+πn
x={-7π/2;-8π/3;-5π/2}∈[-7π/2;-5π/2]