Find all solutions in the interval [0,2pi).

cos 2x + sqrt (2) sinx=1

cos 2x + sqrt (2) sinx=1

Note that: cos 2x = cos^2x - sin^2x = (1-sin^2x) - sin^2x = 1 - 2sin^2x.

So, when alternatively written, you have the following equation:

- 2sin^2x + sqrt (2) sinx + 1 = 1

- 2sin^2x + sqrt (2) sinx = 0

Then, let z=sin (x). So you get,

- 2z^2 + sqrt (2) z = 0

z ( - 2z + sqrt (2)) = 0

Either z=0, or - 2z + sqrt (2) = 0 - - - > z=sqrt (2) / 2.

Then, since z=0 or z=sqrt (2) / 2, therefore sin (x) = 0, or sin (x) = sqrt (2) / 2.

Then, for you remains just to list the angles. (Let me know if this is not fair or if you got questions.)