How do I solve a quadratic trigonometric equation with unusual terms?

[db1uover]

Homework Statement

2 cos x + tan x = sec x

Homework Equations

I can move terms around with identities, but I'm stuck with the partially solved equation below. I don't know how to solve a quadratic with weird terms. I got really far. But how do I show sin x?

sin x = (-1 +- sqrt (1+9))/-4 = 1, -.5 which implies x = pi/2, 7pi/6, 11pi/6 ; range [0, 2pi).

The Attempt at a Solution

cos x (2 cos x + tan x) = sec x (cos x)
2 cos^2 x + sin x = 1
2 cos^2 x - 1 + sin x = 0
(1 - 2 sin^2 x) + sin x = 0
-2sin^2 x + sin x + 1 = 0

 

[rock.freak667]

If you put t=sinx

you will get -2t²+t+1=0

Now use the quadratic equation formula

 

[db1uover]

Thanks Rock. I got confused with the solution given to me. It showed taking the root of 10 gives some kind of rational number. I was so lost. At least I know where I was messing up. Thanks.