You correctly applied the definitions of $\tan$ and $\sec$ to go from the equation in the problem statement to $\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x.$ However, you combined the two fractions incorrectly; in general $\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ For example, you know that $\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$ and that $\frac{1}{4}+\frac{1}{4}\neq\frac{1}{4}.$
The correct form of the equation should be $2\cos^2x-\sin x-1=0.$ You will want to use the fact that for any $x$, $\sin^2(x)+\cos^2(x)=1.$
Thus, any occurrence of $\cos^2(x)$ can be replaced with $(1-\sin^2(x))$. $2(1-\sin^2(x))-\sin(x)-1=0$ $[2\cdot1-2\cdot\sin^2(x)]-\sin(x)-1=0$ $-2\cdot\sin^2(x)+(-1)\sin(x)-1+2=0$ $(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$
Then, you will need to use the quadratic formula to solve for $\sin(x)$. It may help to write $y=\sin(x)$ temporarily, to prevent confusion. Keep in mind that there may be two different possible values of $\sin(x)$ that result.
$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$ $(-2)y^2+(-1)y+(1)=0$ $y=\frac{-(-1)\pm\sqrt{(-1)^2-4(-2)(1)}}{2(-2)}=\frac{1\pm\sqrt{1+8}}{-4}=\frac{1\pm 3}{-4}=\begin{cases}\frac{1+3}{-4}=\frac{4}{-4}=\fbox{$-1$} &\text{ or }\\\\ & \\ \frac{1-3}{-4}=\frac{-2}{-4}=\fbox{$\frac{1}{2}$}.\end{cases}$
Lastly, you need to find those values of $x$ that produce that value (or those values) of $\sin(x)$. Keep in mind that the function $\sin(x)$ is periodic; here is part of its graph to illustrate: 
(and it continues similarly out to infinity in each direction) so that there are always infinitely many values of $x$ that will produce a given value of $\sin(x)$.
For any real number $x$, the values of $\sin(x+2\pi k)$ are all identical for any integer $k$,. There are usually even more values you can plug into $\sin$ that will give the same number (look at the graph for an idea).