Question

2cos^2 theta-2√2cos theta+1 = 0​

Answer 1

Taking theta = alpha (just to make the typing easy for me ^_^")

$2 { \cos }^{2} \alpha - 2 \sqrt{2} \cos\alpha + 1 = 0$

Let Cos a be equal to x,

$2 {x}^{2} - 2 \sqrt{2} x + 1 = 0 \\ \\ 2 {x}^{2} - \sqrt{2} x - \sqrt{2} x + 1 = 0 \\ \\ \sqrt{2} x( \sqrt{2} x - 1) - ( \sqrt{2} x - 1) = 0 \\ \\ ( \sqrt{2} x - 1)( \sqrt{2} x - 1) = 0 \\ \\ {( \sqrt{2}x - 1) }^{2} = 0$


Therefore,
the root is,
$\sqrt{2} x - 1 = 0 \\ \\ x = \frac{1}{ \sqrt{2} }$

which is
$\cos \alpha = \frac{1}{ \sqrt{2} } = \cos \frac{\pi}{4}$

Therefore,
$\alpha = \frac{\pi}{4} \:$

Answer 2

$2 {cos}^{2} x - 2 \sqrt{2} cosx + 1 = 0 \\ 2 {cos}^{2}x \: - \sqrt{2} cosx \: - \sqrt{2} cosx + 1 = 0 \\ { (\sqrt{2}cosx \: + 1) }^{2} = 0 \\ cosx \: = - \frac{1}{ \sqrt{2} } \\ x = 2n\pi \frac{ + }{} {cos}^{ - 1} \frac{ - 1}{ \sqrt{2} }$

HOPE THIS HELPS PLZ MARK AS BRAINLIEST.