Question

resolva a desigualdade cosx+raiz quadrada de 3.senx>raiz quadrada de dois, com 0<x<2pi

Answer 1

$cosx+ \sqrt{3}senx\ \textgreater \ \sqrt{2} ;com,0\ \textless \ x\ \textless \ 2 \pi \\ \\ \frac{1}{2}cosx+ \frac{ \sqrt{3} }{2}senx \ \textgreater \ \frac{ \sqrt{2} }{2} \\ \\ cos \frac{ \pi }{3} cosx +sen \frac{ \pi }{3} senx\ \textgreater \ \frac{ \sqrt{2} }{2} \\ \\ cox(x- \frac{ \pi }{3})\ \textgreater \ \frac{ \sqrt{2} }{2} \\ \\ 2k \pi - \frac{ \pi }{4} \ \textless \ x- \frac{ \pi }{3} \ \textless \ \frac{ \pi }{4} +2k \pi \\ \\ 2k \pi - \frac{ \pi }{4}+ \frac{ \pi }{3} \ \textless \ x\ \textless \ \frac{ \pi }{4}+ \frac{ \pi }{3}+2k \pi \\ \\ 2k \pi+ \frac{ \pi }{12}\ \textless \ x\ \textless \ \frac{7 \pi }{12}+2k \pi$
p/k=0 => $\frac{ \pi }{12} \ \textless \ x\ \textless \ \frac{7 \pi }{12}$