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

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SUMMARY

The discussion focuses on solving the quadratic trigonometric equation 2 cos x + tan x = sec x. The user successfully transformed the equation into a quadratic form involving sin x, leading to the equation -2sin² x + sin x + 1 = 0. By substituting t = sin x, they arrived at the quadratic equation -2t² + t + 1 = 0. The solution process involves applying the quadratic formula to find the roots, which include values that correspond to specific angles within the range [0, 2π).

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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
 
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If you put t=sinx

you will get -2t2+t+1=0

Now use the quadratic equation formula
 
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.
 

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