Question

а) cos ^2 ( х + п/3) = 1

б) sin^2 ( 2х - п/6 ) = 3/4

Answer 1

$sin^2(2x-fracpi6)=frac34;\ sin (2x-fracpi6)=pmfrac{sqrt3}{2};\ 2x-fracpi6=pm(-1)^ncdotfracpi3+pi n;\ 2x=fracpi6pm(-1)^ncdotfracpi3+pi n;\ x=fracpi{12}pm(-1)^ncdotfracpi6+frac{pi n}{2};$a)
$cos^2(x+fracpi3)=1;\ cos(x+fracpi3)=pm1;\ x+fracpi3=pi n, nin Z;\ x=-fracpi3+pi n, nin Z;$
б)