General Solution of $\sin\theta=3\cos\theta$ I'm a high school level maths student currently working through some exercises for the general solution of trigonometric equations and have come across this one that I am stuck on. Any hints would be much appreciated! 
Question:

Determine the general solution of the trigonometric equation:
  $\sin\theta=3\cos\theta$

So the first thing I think of doing when seeing this is making sure either side has the same ratio. I can do this using co-ratios, so:
$\sin\theta=3\sin(90^\circ-\theta)$
$\sin\theta=3\cos\theta$
From here I get stuck; there are no values to compute a reference angle with.
The textbook gives the solution as:
$\theta=71.6^\circ+k\cdot 180^\circ, k\in\mathbb{Z}$
Can anyone give me a hint? Thanks in advance!
-Shaun 
 A: (I assume you know how to use the inverse trigonometric functions of a calculator.)  Divide both sides by $\cos\theta$.  What do you get?
A: Dividing both sides by $\cos \theta$ yields $$\frac{\sin \theta}{\cos \theta} = 3\frac{\cos \theta}{\cos \theta}$$ but we know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ so we get $$\tan \theta = 3$$
Hence $$\theta = \arctan 3 + n\pi, \quad \text{for some integer n}.$$
Or, if you want $$\theta = \arctan 3 + n\cdot 180^{\circ}, \quad \text{for some integer n}.$$
A: Notice, 
$$\sin\theta=3\cos\theta $$ $$\implies 3\cos\theta-\sin\theta=0 $$
Now, divide the above equation by $\sqrt{3^2+(-1)^2}=\sqrt{10}$
$$\frac{3}{\sqrt{10}}\cos\theta-\frac{1}{\sqrt{10}}\sin\theta=0 $$
Now, let $\frac{3}{\sqrt{10}}=\cos\alpha \implies \sin\alpha=\frac{1}{\sqrt{10}}$, we have $$\cos\theta\cos\alpha-\sin\theta\sin\alpha=0  $$
$$\implies \cos(\theta+\alpha)=0  $$ Now, writing the general solution, we get $$\theta+\alpha=(2n+1)\frac{\pi}{2}$$ 
$$\implies \theta=(2n+1)\frac{\pi}{2}-\alpha$$ Substituting the value of $\alpha$
$$\implies \color{blue}{\theta=(2n+1)\frac{\pi}{2}-\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)}$$
$$\text{Or} \quad \color{blue}{\theta=(2n+1)\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)}$$
Where, $\color{blue}{\text{n is any integer} }$
A: $$\sin \theta=3\cos \theta$$
$$\frac{\sin \theta}{\cos \theta}=3$$
$$\tan \theta=3$$
$$\theta=\tan^{-1} (3)+180^\circ k$$
$$\theta\approx71.56505118^{\,\circ}+180^\circ k $$
