How do you solve $cos2x=[sqrt(2)/2]$ over the interval 0 to 2pi?

Question

24332 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T20:37:57

$S={pi/8, (7pi)/8, (9pi)/8,(15pi)/8}$

$2x=cos^-1 (sqrt 2/2)$

$2x=+-pi/4 + 2pin$

$x=+- pi/8 + pi n$

$n=0, x=pi/8, -pi/8$

$n=1, x=(9pi)/8, (7pi)/8$

$n=2, x=(17pi)/8, (15pi)/8$

$S={pi/8,(7pi)/8, (9pi)/8,(15pi)/8}$

How do you solve cos2x=[sqrt(2)/2]cos2x=[sqrt(2)/2] over the interval 0 to 2pi?