simplify the rational trigonometric function. if the denom cos²x - cos x - 2 / 6cos x - 12 numerator =…

Explanation:

Step1: Factor the numerator

Let $t = \cos x$. Then $\cos^{2}x-\cos x - 2=t^{2}-t - 2$. Using the formula for factoring quadratic expressions $ax^{2}+bx + c=a(x - x_1)(x - x_2)$ where for $t^{2}-t - 2$, $a = 1$, $b=-1$, $c = - 2$ and the roots of $t^{2}-t - 2=0$ are $t_1=\frac{1+\sqrt{1+8}}{2}=2$ and $t_2=\frac{1 - \sqrt{1 + 8}}{2}=-1$. So $t^{2}-t - 2=(t - 2)(t+1)=\left(\cos x - 2\right)\left(\cos x+1\right)$.

Step2: Factor the denominator

Factor out the greatest - common factor from $6\cos x-12$. The GCF of $6\cos x$ and $-12$ is 6. So $6\cos x-12 = 6(\cos x - 2)$.

Step3: Simplify the rational function

$\frac{\cos^{2}x-\cos x - 2}{6\cos x-12}=\frac{(\cos x - 2)(\cos x + 1)}{6(\cos x - 2)}$. Since $\cos x\neq2$ (because $-1\leqslant\cos x\leqslant1$), we can cancel out the non - zero factor $\cos x - 2$. The simplified function is $\frac{\cos x + 1}{6}$.

Answer:

$\frac{\cos x + 1}{6}$