# Thread: Missed 3 question on my practice exam

2. ## Re: Missed 3 question on my practice exam

For #1, if you want $\displaystyle y = \cos(x) - \cos\left(\dfrac{x}{2}\right)$ to be zero, then $\displaystyle \cos x = \cos\left(\dfrac{x}{2} \right)$. So, $\displaystyle x = \dfrac{x}{2}$ or $\displaystyle x = \pi - \dfrac{x}{2}$ (by the symmetry of $\displaystyle \cos x$). For the first one, you have $\displaystyle x = 0$. For the second, you have $\displaystyle x = \dfrac{2\pi}{3}$.

For #2, since $\displaystyle \cot \theta = \dfrac{\cos \theta}{\sin \theta}$, it is undefined only when $\displaystyle \sin \theta = 0$. It does not matter what the interval is. $\displaystyle \sin \theta = 0$ when $\displaystyle \cot \theta$ is undefined.

For #3, take log base 5 of both sides to get:
$\displaystyle x^2-4x-31 = 1$ so $\displaystyle x^2-4x-32 = 0$
$\displaystyle (x-8)(x+4) = 0$
$\displaystyle x=-4,x=8$
Hence, $\displaystyle |a+b| = |{-4}+8| = 4$

3. ## Re: Missed 3 question on my practice exam

(1) $\displaystyle 0 \le x \le 2\pi$

let $\displaystyle u = \frac{x}{2} \implies 0 \le u \le \pi$

$\displaystyle \cos(2u) - \cos{u} = 0$

$\displaystyle 2\cos^2{u} - \cos{u} - 1 = 0$

$\displaystyle (2\cos{u} + 1)(\cos{u} - 1) = 0$

$\displaystyle \cos{u} = -\frac{1}{2} \implies u = \frac{2\pi}{3} \implies x = \frac{4\pi}{3}$

$\displaystyle \cos{u} = 1 \implies u = 0 \implies x = 0$

4. ## Re: Missed 3 question on my practice exam

Oops! Thank you skeeter. You are correct. I meant $\displaystyle x=\dfrac{x}{2}$ or $\displaystyle x = 2\pi - \dfrac{x}{2}$, which would have given me the correct answer.

5. ## Re: Missed 3 question on my practice exam Originally Posted by SlipEternal Oops!
I've stated that word more than a few times on this site ... _ _ it happens. #### Search Tags

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