Question

5 - 4sin^2x = 4 cosx

Answer 1

$5 - 4sin^2x = 4 cosx$
$5-4(1-cos^2x)-4cosx=0$
$5-4+4cos^2x-4cosx=0$
$4cos^2x-4cosx+1=0$
$cosx=t; t e [-1;1]$
$4t^2-4t+1=0$
$(2t-1)^2=0$
$t=\frac{1}2$
$cosx=\frac{1}2$
$x=+-\frac{ \pi }3+2 \pi n; n e Z$