![\bf cos(2x)=-\cfrac{\sqrt{2}}{2}\implies cos^{-1}[cos(2x)]=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right) \\\\\\ \measuredangle 2x=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)\implies \measuredangle 2x= \begin{cases} \frac{3\pi }{4}\\\\ \frac{5\pi }{4} \end{cases}\\\\ -------------------------------\\\\ x=\cfrac{3\pi }{2\cdot 4}\implies \measuredangle x=\cfrac{3\pi }{8}\\\\ -------------------------------\\\\ x=\cfrac{5\pi }{2\cdot 4}\implies \measuredangle x=\cfrac{5\pi }{8}](/tpl/images/0461/4656/b2e70.png)
Precalc.
problem attached.
a full explanation will be rewarded.
...
Mathematics, 19.01.2020 04:31 julesperez22
Precalc.
problem attached.
a full explanation will be rewarded.
Answers: 3
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