2sinx-cosx=3/sqrt. of 2

Answer 1

2sinx = 3/√2 + cosx
4sin^2 x = 9/2 + 3√2 cosx + cos^2 x
4 - 4cos^2 x = 9/2 + 3√2 cosx + cos^2 x
5cos^2 x + 3√2 cosx + 1/2 = 0

cosx = [-3√2 ± √(18-10)]/10
= (-3√2 ± 2√2)/10
= -√2/10 or -1/√2

now just take arccos of those values.
Obviously x = 3π/4 is one, but I dunno the other.