Решите................

Решите задачу:

$(16^\sin x)^\cos x+\frac{6}{4^{\sin^2\left(x-\frac{\pi}{4}\right)}}}=4;\\ 16^{\sin x\cdot\cos x}+\frac{6}{4^{\frac{1-\cos\left(2x-\frac{\pi}{2}\right)}{2}}}}=4;\\ 4^{\sin 2x}+\frac{6\cdot 2^{\cos\left(2x-\frac{\pi}{2}\right)}}{2}}=4;\\ 2^{2\sin 2x}+3\cdot 2^{\sin 2x}=4;\\ 2^{\sin 2x}=t\ \textgreater \ 0 \Rightarrow t^2+3t-4=0;\, t_1=-4,\,t_2=1.\\ t=1\Rightarrow 2^{\sin 2x}=1\, \Rightarrow \sin 2x=0;\\ 2x=\pi n,\, n\in \mathbb{Z},\\ x=\frac{\pi n}{2},\, n\in \mathbb{Z}.$