Help With Double Angles And Trig Identity Problem Hello please help me with these trig identities and double angles as I am not sure where I am going wrong but I keep getting the wrong answer 
This is the problem
$$
\sin(\theta+30) = 2\cos(\theta)
$$
This is my one of my incorrect solutions
$$\sin(\theta +30) = 2\cos(\theta)$$
$$\sin(\theta)\cos(30) + \sin(30)\cos(\theta)=2(1 - \sin(\theta))$$
$$\sin(\theta)(\frac{\sqrt3}{2})+(\frac{1}{2})(1 - \sin(\theta))=2-2\sin(\theta)$$
$$\frac{\sin(\theta)\sqrt3+1-\sin(\theta)}{2}+2 \sin(\theta)=2$$
I get stuck and I am not sure what to do with this problem.
Please help as I am trying to self teach my A -level maths.
Thanks in advance
 A: 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.
==================

==================
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 $\tan \theta = \sqrt 3.$
==================

================== 
Doing these graphs yourself really does help. It's true.
A: Your addition identity is almost correct. You can't say $\cos(\theta) = 1 - \sin(\theta)$. However, you can say $\cos^2(\theta) = 1 - \sin^2(\theta)$ from the Pythagorean Identity. Just be aware that you CANNOT drop the squares in this equation by taking the square root of both sides. Now,
\begin{array}{ccc}
   \sin(\theta + 30) & = & \sin(\theta)\cos(30) + \cos(\theta)\sin(30) \\
   & = & \sin(\theta)\frac{\sqrt{3}}{2} + \cos(\theta)\frac{1}{2} \\
   & = & \frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right] 
\end{array}
If this expression was equal to $2\cos(\theta)$, we would have:
\begin{array}{ccc}
   \frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right] & = &2\cos(\theta) \\
    \sqrt{3}\sin(\theta) + \cos(\theta)& = & 4\cos(\theta) \\
     \sqrt{3}\sin(\theta) & = &  3\cos(\theta) \\ 
\sin(\theta) & = & \sqrt{3}\cos(\theta)\\
\end{array}
Since this doesn't hold for all $\theta$, this is not an identity. As mentioned in the comments, you can get to $\tan(\theta)$ from what I ended up with. Then, you can solve for $\theta$.
A: What you have written cannot be an identity. If it were then $\sin(\theta + 30)$ must equal $2\cos\theta$ for all values of $\theta$. However, while $\sin(\theta+30)$ oscillates between $-1$ and $1$, we see that $2\cos\theta$ oscillates between $-2$ and $2$. They have different ranges and so cannot possibly be identical functions. Let's assume you want to find particular values of $\theta$ for which $\sin(\theta + 30) = 2\cos\theta.$
The double angle formula tells us that $\sin(\theta + 30) = \sin\theta\cos(30) + \sin(30)\cos\theta.$ Two well-known values of sine and cosine are $\cos(30^{\circ}) = \sqrt{3}/2$ and $\sin(30^{\circ}) = 1/2.$ Thus:
$$\sin(\theta + 30) = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta. $$
It follows that $\sin(\theta+30) = 2\cos\theta$ if and only if
$$\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta = 2\cos\theta \iff \frac{\sqrt{3}}{2}\sin\theta - \frac{3}{2}\cos\theta = 0 \, . $$
Consider the numerator: $\sqrt{3}\sin\theta - 3\cos\theta = 0 \iff \tan\theta = \sqrt{3} \iff \theta = 60^{\circ} + 180n^{\circ},$ where $n$ is any integer. The multiples of $180^{\circ}$ are added because $\tan\theta$ is periodic with a period of $180^{\circ}$; it repeats itself every $180^{\circ}$. The finally answer is then:
$$\theta = \ldots, -120^{\circ}, \ 60^{\circ}, \ 240^{\circ},\ldots $$
$$ \theta \in \{ (60 + 180n)^{\circ} : n \in \mathbb{Z} \}.$$
