7.169. $\left(16^{\sin x}\right)^{\cos x}+\frac{6}{4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}}-4=0$.
Solution.

Transform the denominator of the second term of the equation:

$4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}=4^{\left(\sin x \cos \frac{\pi}{4}-\cos x \sin \frac{\pi}{4}\right)^{2}}=4^{\left(\frac{\sqrt{2}}{2}\right)^{2}\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)}=4^{\frac{1}{2}(1-\sin 2 x)}=$
$=4^{\frac{1}{2}-\frac{1}{2} \sin 2 x}=\frac{2}{2^{\sin 2 x}}$, from which $\frac{6}{4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}}=3 \cdot 2^{\sin 2 x}$. We obtain the equation $\left(2^{\sin 2 x}\right)^{2}+3 \cdot 2^{\sin 2 x}-4=0 \Rightarrow 2^{\sin 2 x}=-4$ (no solutions) or $2^{\sin 2 x}=1$, from which $\sin 2 x=0, x=\frac{\pi n}{2}$, where $n \in Z$.

Answer: $\frac{\pi n}{2} ; n \in Z$.