0 \\ 2cosx+ \sqrt{3} =0 \\ cosx=- \frac{ \sqrt{3} }{2} \\ x=+- \frac{5 \pi }{6} + \pi n \\ 3cosx+4=0 \\ cosx \neq - \frac{4}{3} " alt="(2cosx+ \sqrt{3} )(3cosx+4)=0 \\ tgx>0 \\ 2cosx+ \sqrt{3} =0 \\ cosx=- \frac{ \sqrt{3} }{2} \\ x=+- \frac{5 \pi }{6} + \pi n \\ 3cosx+4=0 \\ cosx \neq - \frac{4}{3} " align="absmiddle" class="latex-formula">
Ответ: х=+- 5π/6+πn, n∈Z