log10(основание) (sin2x-cosx+100) = 2

Question

log10(основание) (sin2x-cosx+100) = 2

Answer 1

log[10](sin2x-cosx+100) = 2 =>  sin2x-cosx+100 = 100

2 sinx cosx-cosx=0

2sinx=cosx/cosx

sinx=1/2

x=(1)^k П/6 + Пk, k принадлежит Z