Question

Найдите cos (x) ... Остальное во вложении.

Answer 1

  
 $sinx+3cosx=2\ sqrt{1-cos^2x}+3cosx=2\ cosx=t\ sqrt{1-t^2}+3t=2\ 1-t^2=(2-3t)^2\ t leq 2/3 \ 1-t^2=4-12t+9t^2\ 10t^2-12t+3=0\ D=144-4*3*10 = sqr{24}^2\ t=frac{12+2sqrt{6}}{20} =frac{6+sqrt{6}}{10}\ t=frac{12-2sqrt{6}}{20} = frac{6-sqrt{6}}{10}\\ cosx=frac{6-sqrt{6}}{10}$