Let's start by copying the equation:
[tex]-5\cos ^2(x)+4\cos (x)+1=0[/tex]To make it easier to see, let's substitute cos(x) by "u":
[tex]-5u^2+4u+1=0[/tex]To find the values of "u", we can use Bhaskara's Equation:
[tex]\begin{gathered} u=\frac{-4\pm\sqrt[]{4^2-4\cdot(-5)\cdot1}}{2\cdot(-5)} \\ u=\frac{-4\pm\sqrt[]{16+20}}{-10} \\ u=\frac{-4\pm\sqrt[]{36}}{-10} \\ u=\frac{-4\pm6}{-10} \end{gathered}[/tex][tex]\begin{gathered} u_1=\frac{-4+6}{-10}=\frac{2}{-10}=-0.2 \\ u_2=\frac{-4-6}{-10}=\frac{-10}{-10}=1 \end{gathered}[/tex]Now, let's substitute cos(x) back:
[tex]\begin{gathered} \cos (x_1)=-0.2 \\ \cos (x_2)=1 \end{gathered}[/tex]Since it is a trigonometric solution, we have repeating values of "x" that satisfy each equation above.
The first, the one you already got, comes from
[tex]\begin{gathered} \cos (x)=1 \\ x=0+2\pi k \\ x=2\pi k \end{gathered}[/tex]The smallest non negative is for k = 0 which gives
[tex]x=0[/tex]The next following this part would be for k = 1, which gives:
[tex]x=2\pi[/tex]However, we have another equation for solutions:
[tex]\cos (x)=-0.2_{}[/tex]For this equation, the smallest "x" value can be found using arc-cossine of -0.2 in a calculator, which gives:
[tex]\begin{gathered} x=\arccos (-0.2) \\ x=1.772\ldots \end{gathered}[/tex]This is the next non-negative solution for the equation, because it is smaller than the other we found.
So the second part is x = 1.772.
