Given $y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$ find the domain of the function

Question

Find the domain of the function $$y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$$

I'm unable to find the values for which $\cos(x)\geq 1/2$.

PS: This is not a homework question.

Answer 1

$\begin{cases}\cos(x)-1/2\ge 0\\6+35x-6x^2>0\end{cases}\iff\begin{cases}\cos(x)\ge1/2 \text{ (*)}\\6x^2-35x-6<0\text{ (**)}\end{cases}$

From $(*)$ we have to find the value $x\in \mathbb R$ such that $\cos(x)\ge1/2\implies-\dfrac{\pi}{3}+2k\pi\le x \le\dfrac{\pi}{3}+2k\pi$, $k\in \mathbb Z$.
Note: for $\alpha\in[0,2\pi]$, $\cos(\pi/3)=\cos(5\pi/3)=1/2 \implies$ if we want to solve the inequality $\cos(x)>1/2$ we get $\pi/3<\alpha<5\pi/3$. Since the $\cos(x)$ function is periodic you have to consider all the other values in $\mathbb R$ which satisfies the condition.

From $(**)$ we have to solve $6x^2-35x-6<0\iff \dfrac{35-37}{12}<x<\dfrac{35+37}{12}\iff-\dfrac{1}{6}<x<6$.
In conclusion $dom:=\Big[-\dfrac{\pi}{3}+2k\pi,\dfrac{\pi}{3}+2k\pi\Big]\cap\Big(-\dfrac{1}{6},6\Big)=\left(-\dfrac{1}{6}, \dfrac{\pi }{3}\right]\cup \left[\dfrac{5 \pi }{3},6\right)$

Answer 2

$$ \begin{cases} \cos x-\frac{1}{2}\ge 0\\ -6 x^2+35 x+6 > 0\\ \end{cases} $$ The second inequality is quite simple. The solution is $-\frac{1}{6}<x<6$

For the first inequality $\cos x \ge \frac12$ see the graph below

$$-\frac{\pi}{3}\le x\le \frac{\pi}{3}\lor \frac{5\pi}{3}\le x\le 2\pi$$ The intersection of the two intervals is $$-\frac{1}{6}<x\leq \frac{\pi }{3}\lor \frac{5 \pi }{3}\leq x<6$$ with interval notation $$\left(-\frac{1}{6}, \frac{\pi }{3}\right]\bigcup \left[\frac{5 \pi }{3},6\right)$$