Question

сколько корней на отрезке [0;4п] имеет уравнение cos2x/(√2/2 + sinx) =0

Answer 1

sinx≠-√2/2⇒x≠(-1)^n+1*π/4+πn
cos2x=0
2x=π/2+πn⇒x=π/4+πn/2 U x≠(-1)^n+1*π/4+πn

x={π/4;3π/4;9π/4;5π/2}∈[0;4π]