I solved this problem by converting $\sin x,\cos x,\sec x$ all in terms of $\tan(x/2)$ which gives a biquadratic.
I found $2$ roots of the biquadratic by hit and trial i.e $\tan^2(x/2) = 1/3$ but the calculations were really lengthy.
I had also tried opening all brackets but I was still unsuccessful in finding some nice way to solve the above equation.
It would be really appreciated if someone gives a nice solution to the above problem.
Thank you.
Observe that (see details below) the LHS is equal to
$$(\sin x + \cos x + 2) ( 2 - \sec x) $$
The first term is always positive, hence the only solutions arise from $\sec x = 2$.
Note: To be fair, I worked backwards from OP's observation that the solutions are from $ \tan^2 \frac{x}{2} = 1/3 \Rightarrow \cos x = 1/2$. By dividing throughout by $(2\cos x -1)$, I got the other term as $ (\tan x + 1 + 2 \sec x)$. The above presentation made it clearer why there were no solutions from this term.
Otherwise, I wouldn't have trusted that there was a possible factorization, and the factorization wouldn't have jumped out at me.
This approach doesn't generalize to an arbitrary constant. It was specially selected to make things work out.
By opening the parenthesis and regrouping: $$2 \sin x +1-2\sin x \tan x -\tan x +2=0$$ Since $\cos x \ne 0$, multiply by it both sides: $$ 2\sin x \cos x -2\sin^2 x -\sin x +3\cos x=0$$ $$ 2\sin x \cos x -\sin x-2+2\cos^2 x +3\cos x=0$$ $$ \sin x(2 \cos x -1)+\cos x(2\cos x-1)+2(2\cos x -1)=0$$ $$ (2\cos x -1)(\sin x+\cos x+2)=0 \implies \cos x= \frac{1}{2}$$
HINT: Next step from OP
$$ \cos x = \frac{1 -\frac13}{1 +\frac13}=\frac12 \to x=\pm\frac{\pi}{3}~ etc. $$
