Verified answer

Zero product principle. Set each factor equal to zero and solve the two equations.

(√2 cos (x) + 1)(√2 sin (x) - 1) = 0

EITHER: (√2 cos (x) + 1) = 0

√2 cos (x) = -1

cos(x) = -1/√2

x = 3pi/4, 5pi/4

OR:

(√2 sin (x) - 1) = 0

√2 sin (x) = 1

sin (x) = 1/√2

x = pi/4, 3pi/4

Three solutions on [0, 2pi)

x = pi/4, 3pi/4, 5pi/4

NOTE that: pi/3 is NOT a solution

(√2 cos(x) + 1)*(√2sin(x) - 1) = 0

==> cos(x) = -1/√2 or sin(x) = 1/√2

==> x = 3pi/4 or 5pi/4 or x = pi/4 or 5pi/4

1/√2 is a standard value when x is related to pi/4

(cosx + cosy = 1.2)^2 cos^2x + 2cosxcosy + cos^2y = 1.44 (sinx + siny = 1.4)^2 sin^2x + 2sinxsiny + sin^2y = 1.96 When you add those equations together you get: 2 + 2cosxcosy + 2sinxsiny = 3.40 cosxcosy + sinxsiny = 1.40/2 cos(x - y) = .7 cos(x - y) = [e^i(x - y) + e^-i(x -y)]/2 e^[i(x - y)] +e^-[i(x -y)] = 1.4 e^x^2 - 1.4e^x + 1 = 0 z^2 - 1.4z + 1 = 0 Two imaginary roots results from this, just let z ~ be 0.7. Then x - y = 0.7 and x = 0.7 + y cos(.7) ~ .7 arccos(.7) = 45.57 degrees x - y = 45.57 degrees cosx + cos(x - 45.57) = 1.2 cosx + 0.7cosx + 0.714sinx = 1.2 sinx + sin(x - 45.57) = 1.4 sinx + 0.7sinx - 0.714cosx = 1.4