Question

1,6 cosx + 1, 6 = 0





tg 3x = 1





2cos x + корень из трех = 0

Answer 1

$1.6cos x+1.6=0$
$cos x+1=0$
$cos x=-1$
$x=-pi+2*pi*k$
k є Z
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$tg(3x)=1$
$(3x)=frac{pi}{4}+pi*k$
$x=frac{pi}{12}+frac{pi*k}{3}$
k є Z
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$2cos x+sqrt{3}=0$
$cos x=-frac{sqrt{3}}{2}$
$x=^+_-(pi-frac{pi}{6})+2*pi*k$
$x=^+_-frac{5pi}{6}+2*pi*k$
k є Z