Question

Sin^22x - 2cos2x + 2=0

Answer 1

Answer:

x=πn, n∈Z

Step-by-step explanation:

if sin²2x=1-cos²2x, then

1-cos²2x-2cos2x+2=0; ⇒ cos²2x+2cos2x-3=0; ⇔ (cos2x+3)(cos2x-1)=0;

$\left[\begin{array}{cc}cos2x+3=0\\cos2x-1=0\\\end{array} \ \ \left[\begin{array}{cc}x \ 'does-not-exist'\\cos2x=1\\\end{array} \ \ 2x=2\pi n, n=Z \ x=\pi n, n=Z$