How do I solve these equations:?

Question

sin^2x + 3sinx+2=0 When the domain 0<=x<=360 degree

the 2nd is cos^2 x+2 cos x+1=0 domain 0<=x<=360

Answer 1

Problem 1.
u = sin x
u^2 + 3u + 2 = 0
u = -3/2 +/- 1/2
u = -1, -2
sin x = -1 [because the sine is never -2]
x = 270 degrees

Problem 2.
u = cos x
u^2 + 2 u + 1 = 0
(u+1)(u+1) = 0
u = -1
cos x = -1
x = 180 degrees

Answer 2

sin^2x + 3sinx+2=0
<=> sinx = -1 or sinx = -2( impossible)
sinx = -1 <=> x = 270 degree ( with the initial condition )

cos^2 x+2 cos x+1=0
<=> (cosx+1)^2=0
<=> cosx = -1
<=> x = 180 degree (with the initial condition)

Answer 3

They're both quadratics. Replace the trig function with a variable.
So the first one becomes x^2 + 3x + 2 = 0
so (x + 2)(x + 1) = 0
x = -1, x = -2
sinx = -1, sinx = -2
then just solve those for x

the second do the same thing
and get (x + 1)^2 = 0
so cosx = -1