Trig in the Denominator

In order to have a vertical asymptote, we need the denominator to be $0$ . But since $\cos x$ and $\sin x$ are both less than or equal to $1$ , the only way this could happen is if for some $x$ we had $\cos x = |\sin x| = 1$ ; but this is clearly never the case. So the curve has no vertical asymptotes.

The denominator can be written as $\cos(x)^3-2\cos(x)^2+\cos(x)-2$ since $\sin(x)^2=1-\cos(x)^2$ which is the Pythagorean Identity. Then, let $a = \cos{x}$ . By substitution, the denominator is $a^3-2a^2+a-2$ , which can be factored into $a^2(a-2)+1(a-2) = (a^2+1)(a-2) = (\cos(x)^2+1)(\cos(x)-2)$ . Since $\cos(x)$ is nonreal when $\cos(x)^2+1=0$ and $\cos(x)=2$ has no real solution as $-1<\cos(x)<1$ , there are no values of $x$ that can make the denominator $0$ by the Zero Product Property. As a result, there are $0$ vertical asymptotes for $f(x)$ .

Hint:-

Observe that the denominator can be factorized as follows

$\left(\cos x\right)\left(\cos^2x+1\right)-2\left(2-\sin^2x\right)$

$=\left(\cos x-2\right)\left(2-\sin^2x\right)$

$=-\left(\cos x-2\right)\left(\sin x+\sqrt{2}\right)\left(\sin x-\sqrt{2}\right)$

Let $f(x)=\dfrac {g(x)}{h(x)}$ . Considering the numerator function $g(x) = 3\sin x + 2$ , since $\sin x \in [-1,1]$ , $\implies g(x) \in [-1,5]$ . Therefore, vertical asymptotes occur when the denominator function $h(x) = \cos^3 x + 2\sin^2 x + \cos x - 4 = 0$ . Let consider $h(x)$ as follows:

$\begin{aligned} h(x) & = \cos^3 x + 2{\color{#3D99F6}\sin^2 x} + \cos x - 4 & \small \color{#3D99F6} \text{As }\sin^2 x = 1 - \cos^2 x \\ & = \cos^3 x + 2{\color{#3D99F6}(1-\cos^2 x)} + \cos x - 4 \\ & = \cos^3 x - 2\cos^2 x + \cos x - 2 \\ & = \cos x(\cos x-1)^2 - 2 \end{aligned}$

Since $(\cos x - 1)^2 \ge 0$ , $\cos x(\cos x-1)^2$ is positive or negative when $\cos x$ is positive or negative respectively. Therefore, $h(x)$ is maximum when $\cos x$ is positive. We note that $\max (\cos x) = 1$ , when $x=0, \pm 2\pi$ , and $\max ((cos x-1)^2) = 1$ , when $x = \pm \frac \pi 2, \pm \frac {3\pi}2$ . This implies that $\max(\cos x(\cos x-1)^2) < 1$ , $\implies h(x) < -1$ or $h(x) \ne 0$ for all $x$ . Therefore, there is $\boxed 0$ vertical asymptote.