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![\bf 4cos^4(x)-13cos^2(x)+3=0\impliedby \textit{notice, is really just a quadratic} \\\\\\ 4[~~[cos(x)]^2~~]^2-13[cos(x)]^2+3=0 \\\\\\\ [4cos^2(x)-1][cos^2(x)-3]=0\\\\ -------------------------------\\\\ 4cos^2(x)-1=0\implies 4cos^2(x)=1\implies cos^2(x)=\cfrac{1}{4} \\\\\\ cos(x)=\pm\sqrt{\cfrac{1}{4}}\implies cos(x)=\pm\cfrac{\sqrt{1}}{\sqrt{4}}\implies cos(x)=\pm\cfrac{1}{\sqrt{2}}](/tpl/images/0086/6749/dd8bc.png)
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Solve each equation over [0,2pi)4cos^4x-13cos^2x+3=0...
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