![\bf 2cos^2(4x)-1=0\implies 2cos^2(4x)=1\implies cos^2(4x)=\cfrac{1}{2} \\\\\\ cos(4x)=\pm\sqrt{\cfrac{1}{2}}\implies cos(4x)=\cfrac{1}{\sqrt{2}}\implies cos(4x)=\cfrac{\sqrt{2}}{2} \\\\\\ cos^{-1}[cos(4x)]=cos^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right)\implies 4x = cos^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\\\ 4x = \pm\cfrac{\pi }{4}\implies \boxed{x = \pm\cfrac{\pi }{16}} \ne \cfrac{3\pi }{16}](/tpl/images/0538/5733/c1d78.png)
True or false The value 3pi/16
is a solution for the equation 2 cos^2 (4x)-1 = 0....
Mathematics, 09.03.2020 00:27, sarbjit879
True or false The value 3pi/16
is a solution for the equation 2 cos^2 (4x)-1 = 0.
Answers: 3
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