$pt\iff \sqrt{\frac{1}{16}+cos^4{x}-\frac{1}{cos^2x}}+|cos^2x-\frac{3}{4}|=\frac{1}{2}$.
Đặt $t=cos^2x,t\in [0;1]$
$\implies pt\iff \sqrt{\frac{1}{16}+t^2-\frac{1}{t}}+|t-\frac{3}{4}|=\frac{1}{2}$.
Nếu $t\in [\frac{3}{4};1]\implies \sqrt{\frac{1}{16}+t^2-\frac{1}{t^2}}=\frac{5}{4}-t$
$\iff \frac{1}{16}+t^2-\frac{1}{t^2}=\frac{25}{16}-\frac{5t}{2}+t^2\iff 40t^3+24t^2-16=0\iff t=1$.
$\implies cos(x)=1...v...cos(x)=-1$ (Đến đây dễ rồi nhé).
Nếu $t\in [0;\frac{3}{4})\implies \sqrt{\frac{1}{16}+t^2-\frac{1}{t^2}}=t-\frac{1}{4}$
$\frac{1}{16}+t^2-\frac{1}{t^2}=t^2-\frac{t}{2}+\frac{1}{16}(t \in [\frac{1}{4};\frac{3}{4}))$
$\frac{1}{t^2}=\frac{t}{2}\iff t=\sqrt[3]{2}>1(l)$.
Vậy pt đã cho có nghiệm là: $x=k2\pi...v...x=\pi+k2\pi(k\in Z)$