The question asks for no decimals, but I end up with a solution that contains a decimal, what am I doing wrong?
$$6cos^2(\frac{x}{2}) - 7cos(\frac{x}{2})+2=0 \text{ ; }{\forall x \in \mathbb{R}} \text{ ; NO DECIMALS}$$ $$(3cos(\frac{x}{2}) - 2)(2cos(\frac{x}{2})-1)=0$$ $${cos(\frac{x}{2})=\frac{2}{3}}\text{ , }{cos(\frac{x}{2})=\frac{1}{2}}$$ $$cos^-1(\frac{1}{2})= \frac{\pi}{3}$$ $$cos^-1(\frac{2}{3})=0.841068671$$ $$\frac{\pi}{3} * 2 = \frac{2\pi}{3}$$ $$0.841068671*2 = 1.682137342$$ $$x = \frac{2\pi}{3} + 4\pi k \text{ , k any integer }$$ $$x = 1.682137342 + 4\pi k \text{ , k any integer }$$
is the four pi period correct?
You are expressing the answer in terms of decimals. It intends for you to leave your answer in terms of pi.
$x$ over the interval $[−\infty, \infty]$ satisfies $6\cos^2(x2)−7\cos(x2)+2=0$ for values $x=\frac2{3}(6\pi k\pm \pi)$ and $x = 2(2\pi k\pm\arccos(\frac2{3}))$ for $k \in \mathbb{Z}$