Question

4sinL + 4cosL = 1 Найдите sinL cosL

Answer 1

$4sin L+4cos L=1\\sin L+cos L= dfrac{1}{4} \\(sin L+cos L)^2= bigg(dfrac{1}{4}bigg)^2\\sin^2 L+2sin Lcos L+ cos^2 L= dfrac{1}{16} \\1+2sin Lcos L=dfrac{1}{16} \\sin Lcos L=-dfrac{15}{32}$