Question

Log2(2sin^2 2x + 1) - 2log2cos x = 1 + log2 5

Answer 1

Lg2(2sin^2 2x+1)-log2 cos^2x=lg2 2+lg2 5
(2sin^2 2x+1)/cos^2x=10
2*4sin^2xcos^2x+1=10cos^2x
8cos^2x(1-cos^2x)+1=10cos^2x
8cos^2x-8cos^4x+1=10cos^2x
cos^2x=t  t>0
8t^2+2t-1=0
t=(-1+-3)/8
t1=1/4  t2=-1/2
cosx=1/2  cos=-1/2
x=+-п/3+2пk
x=П+-П/3+2Пk