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.)

The answer mai bc