Solve the equation (0°≤A≤360°)
a) √3sinA-cosA=√2​

This is the guy aboves work, I just wrote it out so you can copy and paste easier.
2sin^2x-cos^2x-2=0
sin^2x+cos^2x=1
cos^2x=1-sin^2x
2sin^2x-(1-sin^2x)-2=0
2sin^2x -1 +sin^2x-2=0
3sin^2x-3=0
3sin^2x=3
sin^2x=1
sin x =(sqr root)1
sin x= + or - 1
sin x =1 sin x = -1
x= sin^-1(1) x=sin^-1 (-1)
x=90+n360 x=270 + 360n
x=90, 270

The range would be 13 because you’re finding the difference