Identity help! precalc

I've been trying to do this, but I'm stuck. try to get it so it's in the form ax^2 + bx + c = 0

sin2x+cosx =0

I used the double angle formula to get
2sinxcosx+cosx=0

but I'm stuck, I'm bad at this stuff, and need it for my HW. Thankyou to whoever helps!

official name for this question: Finding solutions in an interval for an equation with sine and cosine using double-angle identities

2sin(x)cos(x) + cos(x) = 0
cos(x) * (2sin(x) + 1) = 0

cos(x) = 0
x = pi/2 + pi * k

2sin(x) + 1 = 0
2sin(x) = -1
sin(x) = -1/2
x = 7pi/6 + 2pi * k , 11pi/6 + 2pi * k


x = pi/2 + pi * k , 7pi/6 + 2pi * k , 11pi/6 + 2pi * k

or in degrees

x = 90 + 180k , 210 + 360k , 330 + 360k

k is an integer

You didn't place a domain. Watch what happens when we set k = 1

x = pi/2 + pi * k
k = 1
x = pi/2 + pi * 1
x = pi/2 + pi
x = pi/2 + 2pi/2
x = 3pi/2

How'd that happen?!

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