Question

Giải hộ mình pt 2cos8x-8cos4x+5=0 nhanh nha mọi người

Answer 1

2cos8x-8coos4x+5=0

<=>2(2cos²4x-1)-8cos4x+5=0

<=>4cos²4x-8cos4x+3=0

<=>cos4x=\(\dfrac{1}{2}\) <=> 4x=\(\dfrac{\pi}{3}+k2\pi\) hoặc 4x=\(\dfrac{-\pi}{3}+k2\pi\)

<=>x=\(\dfrac{\pi}{12}+\dfrac{k\pi}{2}\) hoặc x=\(\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\)