Please help me to prove that cos^2A =cos4A.pls pls pls

$cos^{2} a = cos^{2}2a - sin^{2}2a \\ \frac{1 + cos2a}{2} = 2cos^{2}2a - 1 \\ 1 + cos2a = 4cos^{2}2a - 2 \\ 4cos^{2}2a - cos2a - 3 = 0 \\ D = 1 - 4*4*(-3) = 49 = 7^{2} \\ cos2a = \frac{1 - 7}{8} = - \frac{3}{4} \\ or \\ cos2a = \frac{1+7}{8} = 1 \\ \\$

$2a = ^+_- arccos (-\frac{3}{4} ) + 2 \pi n$ , n ∈ Z ;
$a = ^+_- \frac{1}{2} arccos (-\frac{3}{4} ) + \pi n$ , n ∈ Z

2a = 2пk, k ∈ Z
a = пk, k ∈ Z

ответ: $a = ^+_- \frac{1}{2} arccos (-\frac{3}{4} ) + \pi n$ , n ∈ Z ;
a = пk, k ∈ Z