Question

Solve for all values of a on the interval [0, 2pi].2 cos²x+3 = -cos x + 4

Answer 1

The Solution:

Given:

`\begin{gathered} 2cos²x+3=-cosx+4 \\ \\ interval:\text{ }[0,2] \end{gathered}`

This becomes:

`\begin{gathered} y=2cos^2x+3+cosx-4=0 \\ \\ y=2cos^2x+cosx-1=0 \end{gathered}`
`Let\text{ }a=\cos x`

So,

`\begin{gathered} 2a^2+a-1=0 \\ \\ 2a^2+2a-a-1=0 \end{gathered}`
`\begin{gathered} 2a(a+1)-1(a+1)=0 \\ \\ (2a-1)(a+1)=0 \\ \\ 2a-1=0\text{ or }a+1=0 \\ a=(1)/(2)\text{ or }a=-1 \\ \\ a=(1)/(2)\text{ or }a=-1 \end{gathered}`

Substitute:

`\begin{gathered} cosx=(1)/(2) \\ \\ x=\cos^(-1)((1)/(2))=(\pi)/(3) \end{gathered}`

Or

`\begin{gathered} cosx=-1 \\ \\ x=\cos^(-1)(-1)=\pi \end{gathered}`

Therefore, the correct answers are:

`x=(\pi)/(3)\text{ or }x=\pi`