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

$\cos x=\frac{-1}{1}$