Question

20 баллов!!Помогите пожалуйста

Answer 1

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

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

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

Ответ. $x=dfrac{pi l}{2},~linmathbb{Z}$

Answer 2

спасибо большое) вы мой спаситель!!!