[tex]\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}
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cos(x)=\pm\sqrt{\cfrac{1}{4}}\implies cos(x)=\pm\cfrac{\sqrt{1}}{\sqrt{4}}\implies cos(x)=\pm\cfrac{1}{\sqrt{2}}[/tex]
[tex]\bf cos(x)=\pm\cfrac{\sqrt{2}}{2}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{4}\\\\
\frac{3\pi }{4}\\\\
\frac{5\pi }{4}\\\\
\frac{7\pi }{4}
\end{cases}\\\\
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cos^2(x)-3=0\implies cos^2(x)=3\implies cos(x)=\pm\sqrt{3}
\\\\\\
cos(x)\approx \pm 1.7[/tex]
now, recall that, cosine for any angle has a range from -1 to 1, anything beyond that, is an invalid value, thus certainly 1.7 is so. Meaning, there's no such angle(s) for the second root.
