Question

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

Answer 1

`\bf f(x)=4cos(2x-\pi )\implies 0=4cos(2x-\pi )\implies 0=cos(2x-\pi ) \\\\\\ cos^(-1)(0)=cos^(-1)[cos(2x-\pi )]\implies cos^(-1)(0)=2x-\pi \\\\\\ 2x-\pi = \begin{cases} (\pi )/(2)\\ (3\pi )/(2) \end{cases}\\\\ -------------------------------\\\\ 2x-\pi =\cfrac{\pi }{2}\implies 2x=\cfrac{3\pi }{2}\implies \measuredangle x=\cfrac{3\pi }{4}\\\\ -------------------------------\\\\ 2x-\pi=\cfrac{3\pi }{2}\implies 2x=\cfrac{5\pi }{2}\implies \measuredangle x=\cfrac{5\pi }{4}`