Question

Giải phương trình : 8cos 4 x – 4cos2x + sin4x – 4 = 0

Answer 1

Answer:

Step-by-step explanation:

$8(1+cos2x/2)^{2} -4c0s2x+sin4x-4=0\\2(1+2cos2x+cos^{2}2x )-4cos2x+sin4x-4=0\\2cos^{2} 2x+sin4x-2=0\\1+cos4x+sin4x-2=0\\cos4x+sin4x=1\\sin(4x+\pi /4)=sin\frac{\pi }{4} \\\left \{ {{4x+\frac{\pi }{4} =\frac{\pi }{4}+k2\pi } \atop {4x+\frac{\pi }{4} =\frac{3\pi }{4} +k2\pi$