Question

Where are the x-intercepts for f(x) = −4cos(x − pi over 2) from x = 0 to x = 2π?

Answer 1

Recall that to get the x-intercepts, we set the f(x) = y = 0, thus

$\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2} \right)\implies 0=cos\left(x-\frac{\pi }{2} \right) \\\\\\ cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2} \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2} \\\\\\ x-\cfrac{\pi }{2}= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}$

$\bf -------------------------------\\\\ x-\cfrac{\pi }{2}=\cfrac{\pi }{2}\implies x=\cfrac{\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{2\pi }{2}\implies \boxed{x=\pi }\\\\ -------------------------------\\\\ x-\cfrac{\pi }{2}=\cfrac{3\pi }{2}\implies x=\cfrac{3\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{4\pi }{2}\implies \boxed{x=2\pi }$