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consider the equation: 2cos^2 (4x)-1=0 final all degree solutions!!! HELP ME!!!!
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The equation is the same thing as cos(8x)=0
keep in mind as @larryboxaplenty pointed out already that \(\bf 2cos^2(\theta)-1\implies cos({\color{red}{ 2}}\theta)\qquad thus\\ \quad \\ 2cos^2(4x)-1\implies cos(({\color{red}{ 2}})4x)\implies cos(8x)\\ \quad \\ 2cos^2(4x)-1=0\implies cos(8x)=0\\ \quad \\ \implies cos^{-1}[cos(8x)]=cos^{-1}(0)\)
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