Mathematics, 24.07.2019 23:00 amberell1234
Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π?
Answers: 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} \frac{\pi }{2}\\ \frac{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}](/tpl/images/0128/9572/21147.png)
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Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π?...
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