As $\cos^2 \theta + \sin^2 \theta = 1,$ I'm afraid $ \cos \theta = \pm \sqrt{1 - \sin^2 \theta} $ which is useless for your purposes. So leave the cosine on you right-hand side as it is, your substitution is just wrong.
Alright, if you do it properly, you get a relationship between $\sin \theta$ and $\cos \theta$ which can be rewritten as specifying a value for $\tan \theta.$
Meanwhile, I am a big fan of drawing graphs. I recommend you get some graph paper or quadrille paper and draw $\theta,y$ axes with, say, $0 \leq \theta \leq 360^\circ$ and then draw your two curves, $y = \sin (\theta + 30^\circ)$ and $y = 2 \cos \theta.$ It will be fairly clear when they cross.
Here is a jpeg of empty axes I just made. Since one axis is degrees and one real numbers, I did not worry about relative scale. I also used negative degrees, that aspect does not really matter, but I can understand if degrees up to 360 would be preferred. Anyway, anyone can email me for a jpeg. Believe me, the practice in drawing these things is invaluable. I guess I will draw this specific graph next.
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Alright, I did the graphs. As you can see, (and confirm yourself once you see the likely candidate points), the graphs cross at $60^\circ$ and $-120^\circ.$ By adding $360^\circ,$ we find that $-120^\circ$ is equivalent to $240^\circ.$ And these are the places where $\tan \theta = \sqrt 3.$
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Doing these graphs yourself really does help. It's true.