Question

Solve:-2cosx² + cosx + 1 = 0

Answer 1

Step 1:

In this question, we are meant to solve the following:

`-2cos^2x\text{ + cos x + 1 = 0}`

Step 2:

The details of the solution are as follows:

`\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}`

Rearranging, we have that:

`\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 =}(-1)/(2) \end{gathered}`

Case 1:

`\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}`

Case 2:

`\begin{gathered} since\text{ cos x = p} \\ and\text{ } \\ p=\text{ }(-1)/(2) \\ Then,\text{ we have that:} \\ cos\text{ x = }(-1)/(2) \\ Taking\text{ cosine inverse of both sides, we have that:} \end{gathered}`
`x\text{ = 120}^0`

CONCLUSION:

The solution to the equation:

`-2\text{ cos}^2x\text{ + cos x + 1 = 0 }`

are:

`\begin{gathered} x\text{ = 0}^0 \\ or \\ x\text{ = 120}^0 \end{gathered}`