Question

I have been stuck on this problem for a while and I have not been able to get a valid solution Can someone help me?

Answer 1

The equation is given as

`\begin{gathered} \cos ^2x+3\cos x+2=0 \\ (\cos x)^2+3\cos x+2=0 \end{gathered}`

Let m = cos x.

Therefore, we have

`m^2+3m+2=0`

Solving the quadratic equation by factorization, we replace +3m with +m and +2m.

Hence

`m^2+m+2m+2=0`

Factorizing, we have

`\begin{gathered} m(m+1)+2(m+1)=0 \\ (m+2)(m+1)=0 \\ \therefore \\ m+2=0,m+1=0 \\ m=-2,-1 \end{gathered}`

Therefore, we have m = -2 or -1

Remember that

`m=\cos x`

Therefore,

`\begin{gathered} \cos x=-1\text{ } \\ or \\ \cos x=-2 \end{gathered}`

Considering the first situation,

`\cos x=-1`

This can be rewritten as

`\cos x=(-1)/(1)`