In the range $[ - \frac{\pi}{2} , \frac{\pi}{2} ]$ , both $\cos \theta$ and $\sec \theta$ are positive functions.

Applying AM-GM, $\frac{8 \cos \theta + 2 \sec \theta}{2} \geq \sqrt{ (8 \cos \theta) (2 \sec \theta) }$ $\implies 8 \cos \theta + 2 \sec \theta \geq 8$ $\implies 8 \cos \theta + 2 \sec \theta -7 \geq 1$

So, there are no real solutions in this range.

P.S. AM-GM inequality follows from the fact that if $(\sqrt{x} - \sqrt{y})^2 \geq 0$ , then $\frac{x+y}{2} \geq \sqrt{xy}$ .

Jahid's is an elegant solution. I'm showing an elementary one.

The given equation is : $8\cos^2 \theta -7\cos \theta +2=0$ . Discriminant of this equation is $7^2-4\times 8\times 2=49-64=-15<0$ . So both the roots of the equation are complex, and the function has no real root

$8\cos\theta+2sec\theta-7=0$

Multiplying everything by $\cos\theta$ ,

$8 \cos^2 \theta - 7 \cos \theta +2 =0$

By quadratic formula,

$\cos\theta = \dfrac{-(-7) \pm \sqrt{(-7)^2-4(8)(2)}}{2(16)} =\dfrac{7 \pm \sqrt{-15}}{16}$

Since both roots are complex, there are no real solutions.

Let $g(\theta) = acos\theta + bsec\theta$

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Let's find minimum value of $g(\theta)$

$g(\theta) = \sqrt{ab}(\sqrt{\frac{a}{b}}cos\theta + \sqrt{\frac{b}{a}}sec\theta)$

$g(\theta) = \sqrt{ab}(\sqrt{\frac{a}{b}}cos\theta + \frac{1}{\sqrt{\frac{a}{b}}cos\theta})$

$g(\theta) = \sqrt{ab}(k + \frac{1}{k}) \text{ where } k = \sqrt{\frac{a}{b}}cos\theta$

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Now, $k + \frac1k \leq -2 \text{ or } k + \frac1k \geq 2$

$\sqrt{ab}(k + \frac1k) \leq -2\sqrt{ab} \text{ or } \sqrt{ab}(k + \frac1k) \geq 2\sqrt{ab}$

$g(\theta) \leq -2\sqrt{ab} \text{ or } g(\theta) \geq 2\sqrt{ab}$

Putting $a = 8$ and $b = 2$

$g(\theta) \leq -8 \text{ or } g(\theta) \geq 8 ....... (1)$

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For roots of $f(\theta) = 8cos\theta + 2sec\theta - 7$

$8cos\theta + 2sec\theta -7 = 0$

$8cos\theta + 2sec\theta = 7$

$g(\theta) = 7$ which is not possible as minimum positive value of $g(\theta)$ is 8 by (1)

Therefore, $\color{#3D99F6}{\boxed{roots = 0}}$