Answer :
SOLUTION:
Step 1:
In this question, we are meant to solve the following:
[tex]-2cos^2x\text{ + cos x + 1 = 0}[/tex]Step 2:
The details of the solution are as follows:
[tex]\begin{gathered} -2\text{ cos}^2x\text{ + cos x + 1 = 0} \\ Let\text{ p = cos x , then we have that:} \\ -2p^2+\text{ p + 1=0} \end{gathered}[/tex]Rearranging, we have that:
[tex]\begin{gathered} 2p^2-p-1=\text{ 0} \\ Factorizing,\text{ we have that:} \\ 2p^2-2p\text{ +p - 1= 0} \\ 2p\text{ }(\text{ p - 1})+1(p\text{ -1})\text{ = 0} \\ (p-1)(2p+\text{ 1})\text{ = 0} \\ either\text{ p -1 = 0 or 2p + 1 = 0} \\ p\text{ = 0+ 1 or 2p = -1} \\ \text{p = 1 or p =}\frac{-1}{2} \end{gathered}[/tex]Case 1:
[tex]\begin{gathered} since\text{ p = cos x} \\ and\text{ p = 1} \\ Then, \\ cos\text{ x = 1} \\ Taking\text{ the cosine inverse of both sides, we have that:} \\ \text{x = }\cos^{-1}(\text{ 1}) \\ \text{x = 0}^0 \end{gathered}[/tex]Case 2:
[tex]\begin{gathered} since\text{ cos x = p} \\ and\text{ } \\ p=\text{ }\frac{-1}{2} \\ Then,\text{ we have that:} \\ cos\text{ x = }\frac{-1}{2} \\ Taking\text{ cosine inverse of both sides, we have that:} \end{gathered}[/tex][tex]x\text{ = 120}^0[/tex]CONCLUSION:
The solution to the equation:
[tex]-2\text{ cos}^2x\text{ + cos x + 1 = 0 }[/tex]are:
[tex]\begin{gathered} x\text{ = 0}^0 \\ or \\ x\text{ = 120}^0 \end{gathered}[/tex]