<s>[INST] Determine the general solution for $2\cos 2x−5\cos x+2=0$ Determine the general solution for $2\cos 2x−5\cos x +2=0$
my answer I got was : $1.05+n\pi, 4.19+n\pi$
 [/INST] First, notice that $\cos(2x) \equiv \cos^2x-\sin^2x$ and so
$$2\cos(2x)-5\cos x + 2 \equiv 2\cos^2x - 2\sin^2x - 5\cos x + 2$$
Then we know that $\sin^2x = 1-\cos^2x$, meaning that
$$2\cos(2x)-5\cos x + 2 \equiv 4\cos^2x - 5\cos x$$
We can factorise to give $(4\cos x - 5)\cos x$. The solutions are then just $\cos x = 0$ since $\cos x = \tfrac{5}{4}$ has no real solutions. Hence $x \in \{\tfrac{\pi}{2}+\pi n : n \in \mathbb{Z}\}.$
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