$4^{\cos2x}+4^{\cos^2x}-3=0$
$\Leftrightarrow 4^{2\cos^2x-1}+4^{\cos^2x}-3=0$
$\Leftrightarrow 4^{2\cos^2x}.\dfrac{1}{4} +4^{\cos^2x}-3=0$
$\Leftrightarrow \dfrac{1}{4}(4^{\cos^2x})^2+4^{\cos^2x}-3=0$
Đặt $t=4^{\cos^2x}$
$\Rightarrow \dfrac{1}{4}t^2+t-3=0$
$\Leftrightarrow t=2$ hoặc $t=-6$
+ Nếu $t=2$:
$4^{\cos^2x}=2$
$\Leftrightarrow \cos^2x=\dfrac{1}{2}$
$\Leftrightarrow \dfrac{1+\cos2x}{2}=\dfrac{1}{2}$
$\Leftrightarrow \cos2x=0$
$\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
+ Nếu $t=-6$:
$4^{\cos^2x}=-6$
Ta có $\cos^2x\ge 0\Rightarrow VT\ge 1$
Do đó phương trình vô nghiệm.
Vậy $x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
