Question

Cos x, если sin x = -0,6 и 3/2п<х <2п

Answer 1

$sinx=-0,6\\ \frac{3 \pi }{2}\ \textless \ x\ \textless \ 2 \pi \\\\ \ cosx=б \sqrt{1-sin^2x}=б \sqrt{1-(-0,6)^2}=б \sqrt{1-0,36}=\\=б \sqrt{0,64}=б0,8\\\\(cosx=б0,8;\frac{3 \pi }{2}\ \textless \ x\ \textless \ 2 \pi ) =\ \textgreater \ cosx=0,8$