Question

20 баллов!!Помогите пожалуйста

Answer 1

$\left(16^{\sin x}\right)^{\cos x}+\dfrac{6}{4^{\sin^2\left(x-\frac{\pi}{4}\right)}}-4=0 \medskip \\ 2^{2\sin 2x}+6\left(4^{-\sin^2\left(x-\frac{\pi}{4}\right)}\right)-4=0 \medskip \\ \left(2^{\sin 2x}\right)^2+6\left(4^{-\left(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}\right)^2}\right)-4=0 \medskip \\ \left(2^{\sin 2x}\right)^2+6\left(4^{-\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}\right)-4=0$

$\left(2^{\sin 2x}\right)^2+6\left(4^{-\frac{1}{2}\left(\sin x-\cos x\right)^2\right)-4=0 \medskip \\ \left(2^{\sin 2x}\right)^2+6\left(4^{-\frac{1}{2}\left(1-\sin 2x\right)\right)-4=0 \medskip \\ \left(2^{\sin 2x}\right)^2+6\left(2^{\sin 2x-1}\right)-4=0 \medskip \\ \left(2^{\sin 2x}\right)^2+3\left(2^{\sin 2x}\right)-4=0 \medskip \\ 2^{\sin 2x}=t \Rightarrow t^2+3t-4=0 \medskip \\ (t+4)(t-1)=0$

$1)~t=-4 \medskip \\ 2^{\sin 2x}=-4 \medskip \\ \left(\forall x\in\mathbb{R}\right)2^{\sin 2x}>0 \Rightarrow \varnothing \medskip \\ 2)~t=1 \medskip \\ 2^{\sin 2x}=2^0 \medskip \\ \sin 2x=0 \medskip \\ 2x=\pi l,~l\in\mathbb{Z} \medskip \\ x=\dfrac{\pi l}{2},~l\in\mathbb{Z}$

Ответ. $x=\dfrac{\pi l}{2},~l\in\mathbb{Z}$