Solve the equation in the indicated domain.
1. 2cos^2 x-5cos x +2=0
x E[0,360)
2. sin^2 x+5 sin x +6=0
x E[0,360)
solve both as quadratics that factor
1.
2cos^2 x - 5cosx + 2 = 0
(2cosx - 1)(cosx - 2) = 0
cosx = 1/2 or cosx = 2
but cosx = 2 is not possible, since -1 ≤ cosx ≤ +1
for cosx = 1/2, since it is positive, x could be in I or IV
x = 60°
or
x = 360-60 = 300°
2.
(sinx+ 2)(sinx + 3) = 0
sinx = -2 or sinx = -3
no solution from either part