log10(основание) (sin2x-cosx+100) = 2
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
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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