We can see that the expression is a quadratic expression of the form:
[tex]5\cos ^2(x)+7\cos (x)+2=0[/tex]Now we can say:
[tex]\cos (x)=\tau[/tex]And use the quadratic equation:
[tex]\begin{gathered} 5\tau^2+7\tau+2=0 \\ \tau_{1,2}=\frac{-7\pm\sqrt[]{7^2-4\cdot5\cdot2}}{2\cdot5}=\frac{-7\pm\sqrt[]{49-40}}{10}=\frac{-7\pm3}{10} \\ \tau_1=\frac{-7+3}{10}=-\frac{4}{10}=-\frac{2}{5} \\ \tau_2=\frac{-7-3}{10}=-\frac{10}{10}=-1 \end{gathered}[/tex]The solutions are the values of x such:
[tex]\begin{gathered} \cos (x)=-1 \\ \cos (x)=-\frac{2}{5} \end{gathered}[/tex]We know that if x = π, cos(x) = -1. Thus, π is a solution.
The other solutions are:
[tex]\cos (x)=-\frac{2}{5}[/tex]And since cos has a period of 2π, the solutions are:
The two other solutions for [0, 2pi) are:
[tex]x=\cos ^{-1}(-\frac{2}{5})\approx1.9823[/tex]And:
[tex]x=2\pi-\cos ^{-1}(-\frac{2}{5})\approx4.3008[/tex]
All the solutions in the interval are:
[tex]x=1.9823,\pi,4.3008^{}[/tex]